. . x is max when Sin rot = 0
x is max where Sin rot is 0 and ro is max.
x is max. when rot = 21t
a 21ta
= 00
2
(21t - 0) = 002
1.2.11 PARALLEL ELECTRIC AND MAGNETIC FIELDS
If e and B are parallel to each other and initial velocity ofthe electron
is zero, the magnetic field exherts no force on the electron. The
resulting motion solely depends upon 'e' ( Fig. 1.13 ).
If the electron is possessing initial velocity v 0 and is along the
magnetic field, then also the effect of magnetic fieldis nil (Since, v oy
is parallel to 8). So resultant motion is due to e only. Then the
equation of motion of the electron is,
vy=voy-at
E.e
Where 'a' is the acceleration of the electron and 'a' = - .
m
23
.......... (1.19)
&
y
f

o x
Fig 1. 13 Parallel
electric and magnetic
fields.
( Negative sign is due to the direction of acceleration which is opposite to that of electric
field ).
dy
dt = Voy - at
· 24
Integrating, y = v oy t - Y2 at
2
.
e.l:
a=-
m
Electronic Devices and Circuits
If initially a component of velocity vox is perpendicular to B, the electron will describe
r'rcular motion. The 'Radius' of the circle is independent of E. And, because of the effect of
electric field E, the velocity along the field changes with time. So the resulting motion is helical
with a pitch that changes with time. i.e., the distance travelled along y-axis increases with each
revolution.
1.3 SIMPLE PROBLEMS INVOLVING ELECTRIC AND MAGNETIC FIELDS ONLY
Problem 1.9
A point source of electrons is situated in mutually perpendicular uniform magnetic and electric
fields. The magnetic flux density is 0.0 I wb/m
2
and the electric field strength is 10
4
V 1m. Determine
the minimum distance from the source at which an electron with zero velocity will again have zero
velocity. .:;._ ,
A
• c
Fig 1.14 For Problem 1.9.
Solution
When the electron is under the influence of perpendicular magnetic and electric fields, its
motion is cycloid. An electron emitted with zero velocity at A, will have again zero velocity at B
(FigI.14).
Therefore, AB is the minimum distance from the source of electrons (A) where the electrons
will have again zero velocity. The expression for x, the distance travelled by the electron is
( See Section 1.2.10 ),
From Eg. ( 1.19), x = -;-[oot - Sin wt]
00
AB = Be = Xmax
a
The expression for Xmax = 2 x 27t
ro
27t
- =t
ro
eV e.l:
a=-=-
m.d m
Electron Dynamics and CRO
I Problem 1.10
Be
(j)= -
m
2n: 2n:m
(.: (j) = Angular frequency = T; T = Be )
e.exm
2
x2n: em 2n:x10
4
x10-
11
-----;;;:----;0-- - - X - x 2n: - ---::---
~ a x = 2 2 - B2 e - ( )'
mxB e 0.01-x1.76
e
.. - = 1.76 x lOll C/Kg
III
25
Two 50-e V electrons enter a magnetic field of 2.0 m wb/m
2
as shown in Fig 1.15, one at 10° and
the other at 20°. How far apart are these electrons when they have travelled one revolution in
their hel ical paths?
V 02
~ ~ - - ~ ~ - - - - - - - - - - - - - - - - - + B
Fig 1.15 For Problem 1.10.
Solution
Since the electron velocities are making angle 8
1
, and 8
2
with the field, the path of the electrons
will be helical.
Energy = 50 eV
Accelerating potentia! V 0= 50 volts.
Velocity,
~
V
o
= V ~ = 5.93 x 10
5
Fa = 5.93 x 10
5
x Eo = 4.2 x 10
6
m/s
35.5xlO-
12
't' = 3 = 1.785 x 10-
8
sec
2x 10-
Period of rotation
Components of velocities along Bare
vI =v
OI
Cos 10°= V
o
(0.9848)
v
2
= v
02
Cos 20° = V
o
(0.9397)
v I - v
2
= Vo (0.9848 - 0.93987) = Vo (0.0451)
Distance = (VI - v
2
) T = velocity x time
= 0.0451(4.2 x 10
6
)(1.785 x 10-
8
)
= 0.338 cm
26 Electronic Devices and Circuits
1.4 PRINCIPLES OF CRT
1.4.1 BASIC CRO CIRCUITRY:
A eRO consists of
1. Vertical amplifier
2. Horizontal amplifier
3. Time base Circuit
4. Power supplies
5. Cathode Ray Tube
The block schematic is shown in Fig. 1.16.
Ext. Horizontal
Output
I I
~ - - - - - - - - ~ 1 1 l
Fig 1.16. Block schematic of Cathode Ray Oscilloscope (CRO).
1.4.2 TYPES OF CRO's
1. Single Beam
2. Double Beam/ Dual Beam
3. Dual trace
4. Storage oscilloscope.
eRO is an extremely useful and versatile laboratory instrument used for the measurement
and analysis of waveforms and other phenomena in electronic circuits. These are basically very
fast x - y plotters.
In the usual eRO application, the x - axis or horizontal input is an internally generated linear
ramp voltage or time base signal. The voltage under examination is applied to the y-axis or as
vertical input of the eRO. When the input voltage is repetitive at fast rate, the display appears as
a stationary pattern on the eRG.
Electron Dynamics and CRO 27
CRO nowadays is being replaced by electro luminescent panels, solid state light emitting
arrays ofGaAs diode and plasma cells. CRT is the heart ofCRO. The operation of CRT may be
described by the five regions mentioned below ( Fig. 1.17 ) :
1. Beam Generating Area The electron beam is generated by
Thermionic Emission
2.
3.
4.
5.
Beam Focus Section
Beam Post Acceleration Region
Beam Deflection Region
Beam Target Region
Grid
Focusing Anode
The beam is focused
Space potential controls the beam
velocity, to enable electrons to reach
screen.
The beam is positioned to the desired
x-y co-ordinates position
Display Screen.
Final Apperture
Cathode
Nickel Cylinder
with a l..ayer of
Barium and Strantium
coated on it
1 """",
c--------------------------------------/
T J--:--
Anode
Accelerating
Anodes
(2000 v -10 Kv )
Vertical
Plate
Ilorizontal
Plate
Fig 1.17 Cathode ~ a y Tube (CRT).
1.4.3 BEAM GENERATION
,
,
Graticules
This is analogous to a Vacuum Device Triode: a cathode supplying electrons, a grid controlling
their rate of emission, and an anode collecting them. In a CRT, there is a small hole or aperture, in
both the grid and anode, permitting a narrow beam to emerge from the anode. The cathode
potential is several thousand volts negative, with respect to anode so that, electrons are liberated
with considerable energy. The Intensity of the beam on the screen can be easily controlled by
varying the grid voltage.
1.4.4 BEAM Focus
It contains the beam focusing electrodes which change the beam pattern as a small round dot.
The two electrodes are focus and astigmatism electrodes. Focus control electrode adjusts the
beam to be concentrated as a dot, where as astigmatism control is used to make the dot as round
as possible.
28 Electronic Devices and Circuits
1.4.5 BEAM DEFLECTIO"l
Many CRTs differ only in this method. Magnetic deflection allows a wider beam deflection angle,
than does electrostatic deflection. Iffull screen beam deflection bandwidth desired is less than 20
KHz, Electromagnetic Deflection System has a substantial cost advantage. Thus TV sets and
medical monitor use Electromagnetic Deflection.
Magnetic Deflection is accomplished by changing magnetic field. This is done by changing
current levels in an inductor. At high frequencies inductors with few turns are necessary to obtain
fast current changes. Inductive reactance increases with frequency.
1.4.6 BEAM POST
This is important for writing speed. The original accelerating field is the potential between cathode
and deflection plates. In electrostatic CRTs, the voltage between cathode and plate is approximately
4 KY. For this a resistance spiral is used inside the tube envelope on which is impressed an
accelerating voltage of 10 KY. The accelerating field bends the beam towards the axis and thus
changes the waveform display or decreases deflection sensitivity.
1.4.7 CRT DISPLAY SCREEN
Phosphor is the usual read out material on the target. It has the capability of converting electrical
energy into light energy. Two phenomena occur when a phosphor is bombarded with a high
energy electron beam. When the beam hits the phosphor, a fluoresence or light emission is observed.
When the excitation beam is removed phosphorescence remains for sometime and indicates where
the phosphor had been stimulated into light emission. The phosphor is classified as
(1) Short persistence (decay in less than 1m sec)
(2) Medium persistence (2 sec)
(3) Long persistence (minutes), human eye tends to peak in 55000 AO,
i.e., yellow to green region. Fluorescent screen material is Zinc Orthosilicate ( P-1
phosphor)
P - 5 screen - Calcium Tungstate gives blue colour.
1.4.8 GR\TICliLES
These are the scale markings on the CRT Screen. They are of three kinds
1. External Graticule
2. Internal Graticule
3. Projected Graticule
External Graticule is screened outside the CRT screen.
The Internal Graticule is screened inside the CRT screen.
Projected Graticule is provided with some covers and allow greater flexibility in
graticule pattern.
Cathode in CRTs is a nickel cylinder with a layer of barium and strontium oxide deposited
over it, to obtain high electron emission at moderate temperatures. Voltages applied to the acceleration
anode vary from 250V to 10,OOOY.
The grid is biased negative with respect to cathode. It controls the density of electrons
being emitted from the cathode. Intensity knob controls the negative voltage of the grid.
Electrol1 DYIl(lmic.\' (lilt! eRO 29
The anode and accelerating anode form electrostatic lens. A typical case is, focusing
;lI1ode is at 1200V, and accelerating anode is at 2000V. Thus a P.D of 800V will produce strong
cb:trostatic field. An electron passing through these anodes has two forces acting on it. The high
accelerating voltage attracts the electrons and speeds it up in a forward direction, and the electrostatic
lield between the t\\'o electrodes tends to deflect the electron. The end result is that all electrons
entering the lens area tend to come together at a point called the focal point.
In CRO focussing is done by varying the focusing electric voltage since the electrons arrive
at the screen v. ith high velocity and there will be secondary emission. These electrons are attracted
0) the acqudag coating and returned to the cathode. Acqudag coating is a Graphite coating.
It is at a po,>itive potential and returns secondary electrons to the cathode.
Dual trace CRO enable the portrayal of two vertical beams. It consists of a single beam
('RT. a time base generator and t\\O identical vertical amplifiers. with an electronic switch.
CI switch alternati\ el)' connects each vertical channel of CRT. after each sweep.
1.4.9 TI!\IE BASE
This permits an operator to display voltage or current variations in time. Many AC signals are
functions of time. So it is an essential feature. Time bases generate an output voltage which is
used to move the beam across CRT screen in a straight horizontallme, and return the beam quickly
back to its starting point. From left to right, the beam moves slov.l)- and so appears as a line. It
returns rapidly from right to left and so cannot be perceived. The left to right or forward movement
of the beam is called trace interval or forward trace.
The sweep action is achieved by a saw tooth wave form, to represent time varying functions.
Time Base Circuit
A typical circuit to deflect electron beam along x-direction on the screen is as shown in Fig. 1.18.
The switch can be a BJT, JFET or any other electronic switching device. When the switch S is
open C gets charged. When S is closed, C discharges. producing a saw tooth wave form. Switching
can be done at a faster rate when electronic devices are used as switches.
v

Fig 1.18 (a) R - C Network. Fig. 1.18 ( b) Time base wave form.
1.5 DEFLECTION SENSITIVITY
1.5.1 ELECTROSTATIC DEFLECTION SENSITIVITY
Electrostatic deflection sensitivity of a pair of deflection plates of a Cathode Ray Oscilloscope
(CRO) is defined as the amount of deflection of electron spot produced when a voltage of I V DC
is applied between the corresponding plates.
30 Electronic Devices and Circuits
Let two plates of length 'I' and spacing's' are kept at a distance D from the screen. Letthe
voltage applied between the plates be V d volts and v the velocity of an electron on entering the
field of the deflection plates (Fig. 1.19).
Then,
Where
I = length of deflecting Plates
s = Spacing between Deflecting Plates
D = Distance between screen and Centre of Deflecting Plates
12 x mv
2
= e x V
v= ~ 2 ~ d
V d voltage of fixed anode in volts
e charge of an electron in coulomb
m = mass of an electron in kgs.
t
I ~
d,
t
f
S __ -r-:.-:'. -:-: •• -:- •• : : • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• d,
tA C G
--------1+-- D,--+I
I ~ D .-1
Fig 1.19 Electrostatic deflection sensitivity.
As the beam passes through the field of the deflection plates, the electrons are attracted
Vd xe
towards the positive plate by a force equal to --
s
Vd xe
F=exE=--
s
The force produces an acceleration of 'a' m/sec
2
and is equal to force divided by mass me
of an electron.
F=ma
F
a= -
m
Vd xe
a=--
sm
The forward motion however continues at velocity v. The time taken by the electron to the
traverse in the field ofthe plates is i sec. The upward velocity attained by the electron in i sec
v v
is v
y
'
I I Vd x e
Then, v = - x a = - x -- m/sec
y v v sm
The ratiQ of upward velocity to the forward velocity at the instant electron leaves the field is
Vy _ Vd xexl
~ - sxme xv
2
Electron Dynamics and CRO
31
The electron follows a parabolic path from A the point of entrance, to point B, the point of
leaving the field. Let the vertical displacement, during this period be d
I
.
Then
d I = ut + YSat
2
d
I
= Y2 X a x ( ~ r
Vd x e x 12
= YS x --"----;:-
sx m X y2
ut = 0
1
t= -
Y
.: upward velocity at t = 0
If e is the angle with the axis that the electron beam makes after emerging out from the field
of the deflection plates. Then
Total deflection,
Let
Then
But
yy d
2
yy
Tan e = - = - or d = D x-
y O
2
2 2 y
d = Vd x ex 1 x O
2
2 s x m xv
2
1
D = D2 + 2"
Vd xexl xD
d = ---"------,-
s x me x y2
y2 = 2xex Va
m
V 10
d=-fl
x
-
Va 2s
for a given CRT, /, D, and s are fixed. Therefore deflection 'd' of the spot can be changed by
changing the ratio Vd .
Va
d 10
SE = Deflection Sensitivity = v;;- = ~ cmN
IxD
............ ( 1.20 )
Deflection Sensitivity SE ( with Electrostatic Field) can be increased by increasing /, but
then the electron may hit the plates. Even after decreasing s, the same problem will be there, We
can increase D, but the size of CRT becomes large. So the only alternative left is by reducing Va'
So a compromise has to be made in the design of CRT, to get optimum value of deflection sensitivity.
32 Electronic Devices and Circuits
1.5.2 ELECTROMAGNETIC DEFLECTION SENSITIVITY
Magnetic Deflection in a Cathode Ray Tube
Electron will get deflected in magnetic field also. So a Cathode Ray Oscilloscope (CRO) can
employ magnetic field as well to get deflective on the screen. Let us derive the expression for the
deflection sensitivity Sm in a magnetic field.
If the magnetic field points out of the paper, the beam is deflected upward. The electron
moves in a straight line from the cathode to the boundaries '0' of the magnetic field. In the
region of the uniform magnetic field, the electron experience a force ofmagnitude e B Vo where Vo
is velocity.
•
p'
Q
p'
8
Jr'------c IA •
...-_____ ---'_.,.-0 .... ·1· .................. f-Z---
Cathode
I ° · :0' 1 p X
1
1
• 1
.( 1
L

Fig 1.20 Magnetic deflection sensitivity.
The path OM will be the arc of a circle whose centre is at Q. The velocity of the particle,
v =
o m
Let,. I = length of the magnetic field
L = Distance between circle of the field and the screen
y = Total y deflection
B = flux density in wb/m
2
Vo = initial velocity
V 0 = accelerating voltage
SM = Deflection sensitivity due to magnetic field
= m/wblm2 = m
3
/wb
The path OM will be the arc of a circle whose centre is at Q.
OM = R8,
where R is the radius of the circle.
since 8 is small. If we assume a small angle of deflection OM::::: I
I
8= -
R
Electron Dynamics and CRO
But radius of the circle is
mv
R=-
eB
33
In all practical cases, L » I, therefore if MP' is projected backwards, it will pass through
the centre 0' of the region of the magnetic field.
y=L tan 8
when 8 is small, y = L8
I l.eB l.e.B
But 8= -=--=--==
R mv
I.B Ie
8 = V;;
I.B Ie
.. y = L V;;
Deflection sensitivity
S -1.
M-
B
Advantages
I I
Compared to electrostatic deflection SM is high, and since SE a -V whereas SM a
o "Vo
Because of high deflection sensitivity, it is used in radars and T.Y. tubes. Radial deflection
can be accomplished easily by rotating coils placed outside, to produce B.
Disadvantages
For high frequency this is not used because of the high reactance produced by the coils.
Ion spot formation creates a black spot at the centre of the screen.
Problem 1.11
In a CRT, the length of the deflecting plates in the direction of the beam is 2 cm, the spacing ofthe
plates is 0.5 cm and the distance of the fluorescent screen from the centre of the plate is 18 cm.
Calculate the deflection sensitivity in m/volt if the final anode voltage is
(a) 500 V (b) 1000 V (c) 1500 V
Solution
ID
Deflection sensitivity = 2sVa
(a) For Va = 500 V
1=2 cm D= 18 cm,
Deflection Sensitivity SE= = 0.072 cmN
(b) For Va = 1000 V, SE = 0.036 cmN.
(c) For Va = 1500 V, SE = 0.024 cmN.
s = 0.5 cm
34 Electronic Devices and Circuits
Problem 1.12
In a CRT, a pair of deflecting plates are 2.0 cm long and are spaced 0.5 cm apart. The distance
from the center of the plates to the screen is 24 cm. The final anode voltage is 1000 V. Calculate
(a) The displacement produced by deflecting voltage of 30V.
(b) The angle which the beam makes with the axis of the tube on emerging from
the field
(c) Velocity of the beam on emerging from the field.
Solution
(a) Deflection produced v y
Vd.l D 30x2x24
d = 2sVa = 2xO.5xlOOO = 1.44cm
d Vd
l
30x2
Tan e = - = -- = = 0.06
D 2sVa 2x1000xO.5
(b)
e = 3
0
, 26'
v
. (c) Tan e =..-1..
-t'-L.:.'--------+ V
v
Resultant Velocity V R' (Fig. 1.21 )
v
v =--
R cose
Fig 1.21 For Problem 1 . 1 ~
~ 2 e V a hr"
v = -- = 5.94 x 105 x V Va = 5.94 x 105 X .JlOOO
me
= 18.75 x 10
6
m/sec
V 18.78x10
6
ve = -co-s-e = ---- = 18.87 x 10
6
mlsec
0.9982
Problem 1.13
The electrons emitted from the thermionic cathode of a Cathode Ray Tube gun are accelerated by
a potential of 400 V. The essential dimensions are L = 19.4 cm, 1= 1.27 cm and d = 0.475cm.
Determine SE. What must be the magnitude of a transverse magnetic field acting over the whole
length of the tube in order to produce the same deflection as that produced by a deflecting potential
of30 V?
Solution
1.D 1.27 x 19.4 x 10-
2
SE = -- = = 0.65 mm/v.
2.sV
a
2 x 0.475 x 400
30 V will produce deflection 000 x 0.98 = 2.67 cm
. L.l.B ~
Y = 2.67 cm = r;rn-. -
...;
2V
o m
L=19.4cm;I=1.27cm;B=? Vo=400V
B = 6.22 x 10-
5
wb/m
2
.
Electro!, Dynamics and CRO 3S
Problem 1.14
Calculate the deflection of a cathode ray beam caused by the earth's magnetic field. Assume the
tube axis is so oriented that it is normal to the field, whose strength is 0.6 G. The anode potential
is 400 V. The Anode - screen distance is 20cm.
o
B
---lr-
y

R
D

Fig 1.22 Deflection in CRT due to Earth's magnetic field.
Solution
The electron starts at 0 ( Fig. 1.22) and is accelerated by anode potential of 400 V. Because of
the effect of earth's magnetic field, it will describe a circle of arc OA, whose radius == AC.
AB is the deflection of the electron on the screen (y).
f2eV::
Now initial velocity, v'
o
= = 1.19 x 10
7
m/sec.
Earth's magnetic field = 0.6 G == 0.6 x 10-4 wb/m
2
Radius of the circle described by the electron due to earth magnetic field
R = 3.37 X 10-
6
ry
B " Yo
= 3.37 x 10-
6
x ../400
6xlO-
5
= I.J 2 m = 112 cm
AC = 112 cm CO = Radius of the circle =AC == 112 cm
CD = (112 - y)
(112 - y)2 + 20
2
= 1122
y2 - 224 Y + 400 == 0 or y = 1.8 cm
36 Electronic Devices and Circuits
1.6 APPLICATION OF CRO
(1) Voltage Measurements
(2) Frequency Measurements
(3) Phase Measurements
(4) T.VDisplay
(5) Indicators Diagrams
(6) Computer Monitors, etc.
1.6.1 CURRENT MEASUREMENT USING CRO
The voltage drop (V) across a resistor 'P' is measured using CRO. The current'!' through the
resistor can be determined using the formula I = V / R.
1.6.2 FREQlJENCY MEASllREMENT
By observing the A.C. signal waveform on CRO, the number of divisions are measured, on the
timescale. Time 'T' is determined by mUltiplying No. of divisions with m.sec (or )..t sec) / Div.
1
Frequency f = T
•
•
•
•
•
•
SUMMARY
V
E = _. F = e x E·,
d '
v= ~ 2 X : V ; y=
Transit Time "t = ~ 2m x d
exV
1 eV = 1.6 x 10-
19
Joules
ex E x t
2
2m
The trajectory of an electron in uniform retarding electric field, when the initial velocity
is making an angle 8 with the field is parabola.
2Vo
X = - x d x Sin 28·
m V '
Force experienced by electron in magnetic field
f = B x I x L Newtons
y IxD
SE = -; = Electrostatic Deflection Sensitivity, SE = 2s Va
SM= Electromagnetic Deflection Sensitivity,
Electron Dynamics and CRO
37
OBJECTIVE TYPE QUESTIONS
1. Electric Field Intensity E = .............................. Its units are ............................. .
2. Expression for the velocity of electron v in terms of acclerating potential
V = ............................. .
3. Force experienced by an electron in a magnetic field of Intensity B,
fm = .............................. Newtons.
4. The acceleration of an electron placed in an electric field of intensity & is
a = ............................. .
5. The trajectory of an electron moving with velocity v in a magnetic field B is
6. The time taken T for one revolution of an electron in a magnetic field, is
T = ............................. .
7. Expression for the radius of an electron placed in a magnetic field of intensity B, and
moving with velocity v is r = ............................. .
8. Expression for electrostatic deflection sensitivity SE = ............................. .
9. Expression for electromagnetic deflection sensitivity SM = ............................. .
10. Graduated scales on the CRO Screen are called as ............................. .
11. Coating which provides the return path for electrons after striking the CRO Screen
12. Cathode Material used in Cathode Ray Tube ............................. .
13. The nature of signal ( wave shape) of Time Base in a CRO is ............................. .
14. Electrostatic deflection sensitivity is defined as ............................. .
15. Electromagnetic deflection sensitivity is defined as ............................. .
16. The energy acquired by an electron in rising through a potential of one volt is
17. A Conductor of length L consists of N electrons. An electron takes T seconds to
travel the distance L. The total number of electrons passing through any cross section
in unit times is .............................. .
18. The trajectory of an electron in two dimensional motion subjected to an electric field
between two plates of a capacitor is .............................. .
19. Relativistic correction should be applied if an electron falls through a potential of
above .............................. volts.
20. When a current carrying conductor oflength L is placed in a magnetic field of strength
B, the force experienced by the conductor is F m = ............................. .
21. When an electron enters the magnetic field with a velocity v in a direction perpendicular
to the magnetic field, the magnetic force results in .............................. motion of the
electron.
22. The .............................. deflection sensitivity is inversely proportional to the square
of accelerating potential.
23. The .............................. deflection is independent of specific charge (elm).
24. When suitable alternating voltages are impressed against two pairs of deflecting p!ates
in a CRO, the famous .............................. figures are obtained on the screen.
25. The mass of a particle moving with a velocity v in terms of its rest mass mo is given
by m = ............................ ..
38
Electronic Devices and Circuits
ESSAY TYPE QUESTIONS
1. Derive the expression for the velocity acquired by an electron, when placed in an
accelerating field &.
2. Derive the expression for radius 'R' and time period 'T' in the case of an electron
placed in a magnetic field of intensity B tesla.
3. Derive the expression for electrostatic deflection sensitivity
4. Derive the expression for electromagnetic deflection sensitivity.
5. With the help of a neat sketch, describe principle and working of Cathode Ray Tube.
6. Give the constructional details of Cathode Ray Tube.
7. Compare Electrostatic and Electromagnetic deflection mechanisms
8. What are the applications of CRO? Give the block schematic of CRO and explain.
MULTIPLE CHOICE QUESTIONS
1. 1 ev = ..... Joules
(a) 1.6 x 10
19
Joules (b) 1.7 x 10-
9
(c) 6.1 x 10-
19
(d) 1.6 x 10-
19
2. If the electron starts at rest with initial velocity = 0 m/Sec and is accelerated by
potential +V, volts the final velocity is,
(a) 5.93)( 10
5
.JV mlsec
(c) 9.53 x 10
5
.JV Km/sec
(b)
(d)
5.9 x 10
5
.JV cmlsec
3.59 x 10
5
.JV mlsec
3. The expression for current I passing through a conductor of cross sectional area
'A'm
1
, in time t secs, with electron density N per m
J
, of charge e, is 1=
Ne Ne Ne NA
(a) -t- (b) t L (c) t A (d) e t
4. The force experienced by an electron in parallel electric field '&' vim and magnetic
field 'B' wb/m
1
, with initial velocity zero m/sec, is ...
Be e. £ B £
(b) m (c) B rrt (d)
Zero
em
5. If aL electron is placed in combined '&' and 'B' fields and if '&' and 'B' are
perpe ldicular to each other, and if the initial velocity is perpendicular to magnetic
field IS, the path of the electron is
(a) Common parabola
(c) Prolate parabola
(b)
(d)
Curtate parabola
Trochoid
6. CRT screens with medium persistance will have visible glow on the screen for a
period of
(a) 1 minute (b) 2 secs (c) 1 m sec (d) few minutes
In this Chapter,
• The basic aspects connected with Semiconductor Physics and the terms
like Effective Mass, Intrinsic, Extrinsic, and Semiconductors are introduced.
• Atomic Structure, Configurations, Concept of Hole, Conductivity inp-type
and n-type semiconductors are given. This forms the basis to study
Semiconductor Devices in the following chapters.
• After p-type and n-type semiconductors, we shall study semiconductor
devices formed using these two types of semiconductors.
• We shall also study the physical phenomena such as corniuction, transport
mechanism, electrical characteristics, and applications of semiconductor
diodes, zener diode, tunnel diode and so on.
40
Electronic Devices and Circuits
In this chapter, we shall first study the basic aspects of Atomic Theory, Electronic structure
of Silicon and Germanium elements, Energy Band Theory, Semiconductor Device Physics, and
then conduction in Semiconductors.
2.1 REVIEW OF SEMICONDUCTOR PHYSICS
2.1.1 ENERGY LEVELS AND ENERGY BANDS
To explain the phenomenon associated with conduction in metals and semiconductors and the
emission of electrons from the surface of a metal, we have to assume that atoms have loosely
bound electrons which can be removed from it.
Rutherford found that atom consists of a nucleus with electrons rotating around it.
The mass of the atom is concentrated in the nucleus. It consist of protons which are positively
charged. In hydrogen atom, there is one positively charged nucleus ( a proton) and a single
electron. The charge on particle is positive and is equal to that of electron. So hydrogen atom is
neutral in charge. The proton in the nucleus carries the charge of the atom, so it is immobile. The
electron will be moving around it in a closed orbit. The force of attraction between electron and
proton follows Coulombs Law. {directly proportional to product of charges and inversely proportional
to (distance )2}.
Assume that the orbit of the electron around nucleus is a circle. We want to calculate the
radius of this circle, in terms of total energy oW' of the electron. The force of attraction between
the electron and the nucleus is :
F a e
2
.
(Therefore, the nucleus has the proton with electron charge equal to 'e')
F a 1
~
( r is the radius of the orbit)
10-
9
where Eo is the permitivity of free space. Its value is -- F/m.
361t
As the electron is moving around the nucleus in a circular orbit with radius r, and with
velocity v, then the force of attraction given by the above expression F should be equal to the
2
centripetal force mv (according to Newton's Second Law of Motion)
r
v
2
( F = m.a; m is the mass of electron and acceleration, a = -)
r
Junction Diode Characteristics
e
2
e
2
The Potential Energy of the electron = ---2 X r = --4--
41tEor 1tEor
( -ve sign is because Potential Energy is by definition work done against the field)
Kinetic Energy = ~ mv
2
But
I e
2
Total Energy W possessed by the electron = -2 mv
2
--4--
1tE
o
r
e
2
mv
2
=--
41tEor
reduces to
e
2
e
2
e
2
W= ----- = ---
81tEor 41tEor 81tE
o
r
e
2
Energy possessed by the electron W = - -8--
1tE
o
r
41
W is the energy ofthe electron. Only for Hydrogen atom W will also be the energy of atom
since it has only one electron. The negative sign arises because the Potential Energy of the
electron is
81tEor
As radius r increases, Potential Energy decreases. When r is infinity, Potential Energy is zero.
Therefore the energy of the electron is negative. If r is < 00, the energy should be less. (Any
quantity less than 0 is negative).
The above equation is derived from the classical model of the electron. But according to
classical laws of electromagnetism, an accelerated charge must radiate energy. Electron is having
2
charge = e. It is moving with velocity v or acceleration ~ around the nucleus. Therefore this
r
electron should also radiate energy. If the charge is performing oscillations with a frequency 'f',
then the frequency of the radiated energy should also be the same. Hence the frequency of the
radiated energy from the electron should be equal to the frequency with the electron orbiting round
the nucleus.
But if the electron is radiating energy, then its total energy OW' must decrease by the amount
equal to the radiation energy. So OW' should go on decreasing to satisfy the equation,
e
2
W=---
81tE
o
r
. . e
2
If W decreases, r should also decrease. Smce If - -- should decrease, this quantity
81tE
o
r
should become more negative. Therefore, r should decrease. So, the electron should describe
smaller and smaller orbit and should finally fall into nucleous. So classical model of atom is not
fairly satisfactory.
42 Electronic Devices and Circuits
2.1.2 THE BOHR ATOM
The above difficulty was resolved by Bohr in 1913. He postulated three fundamental laws.
1. The atom can possess only discrete energies. While in states
corresponding to these discrete energy levels, electron does not emit
radiation in stationary state.
2. When the energy of the electron is changingfrom W
2
to WI then radiation
will be emitted The frequency of radiation is given by
f= W2 -W\
h
where 'h' is Plank s Constant.
h = 6.626 x 10-
34
J - sec.
i.e., when the atom is in stationary state, it does not emit any radiation.
When its energy changes from W
2
to WI then the atom is said to have
moved from one stationary state to the other. The atom remains in the
new state corresponding to WI. Only during transition, will some energy
be radiated.
3. A stationary state is determined by the condition that the angular
momentum of the electron in this state must be an integral multiple of
h127r.
nh
So mvr =-
2tr
where n is an integer, other than zero.
2.1.3 EFFECTIVE MAss
An electron mass 'm' when placed in a crystal lattice, responds to applied field as if it were of
mass m*. The reason for this is the interaction of the electron even within lattice.
E = Kinetic Energy of the Free Electron
p=Momentum
v=Velocity
m = Mass
p=mv
m* = Effective mass of electron
p2
E=-
2m
Electrons in a solid are not free. They move under the combined influence of an external
field plus that of a periodic potential of atom cores in the lattice. An electron moving through the
I ~ t t i c e can be represented by a wave packet of plane waves grouped around the same value of
K which is a wave vector.
Electron velocity falls to zero at each band edge. This is because the electron wave further
)ecomes standing wave at the top and bottom ofa band i.e., Vg = O.
Junction Diode Characteristics 43
Now consider an electronic wave packet moving in a crystal lattice under the influence of
an externally applied uniform electric field. Ifthe electron has an instantaneous velocity Vg and
moves a distance dx in the direction of an accelerating force F, in time dt, it acquires energy dE,
where
5E = F x 5x = F x Vg x 5t
F oE
5E= - x - x 5t
h ok
oE
Vg = hok
Within the limit of small increments in K, we can write,
dK F
dt
.......... ( 2.1 )
But this is not the case for the electron in a solid because the externally applied force is not
the only force acting on the electrons. Forces associated with the periodic lattice are also present.
Acceleration of an electronic wave packet in a solid is equal to the rate of change of its velocity.
d v d (d E ) . d
2
E
Acceleration of an electron in a solid = df = dt dP = (d P )(d t)
But
or
dE
Vg = dp
dk F
dt
h
dk
r = h x-
dt
dk F
dt Ii
~ = (d
P
)
d t d t
from Eq. ( 2.1')
d
2
E F d
2
E
dp2 = t1 . dp2
This is of the form,F = rna, from Newton's Laws of Motion,
where
d
2
E
m* = 1i
2
x --
[ ]
-1
dp2
dv
F=m* x-
g
dV
where m* is the effective mass.
dv
anda= -g
dt
44 Electronic Devices and Circuits
,
If an electric field E is impressed, the electron will accelerate and its velocity and energy will
increase. Hence the electron is said to have positive mass. On the other hand, if an electron is at
the upper end of a band, when the field is applied, its energy will increase and its velocity decreases.
So the electron is said to have negative mass.
In an atom,
e
2
Coulombs force of attraction = 2
41t Eo r
mv
2
Centripetal Force = --
r
Equating these two forces in an atom,
e
2
mv
2
v = velocity; r = radius of Orbit
=--
r
e
2
mv
2
--=--x
41t Eo r
e
2
r = 41tEo mv
2
.......... ( 2.2 )
But
nh
mvr= 21t
h = Plank's Constant; n = Principle Quantum Number.
nh
.. v = 21tmr
Substituting the value ofv in Equation ( 2.2 ),
n
2
h
2
E
r = __ ---'<-0
1tme
2
This is the expression for radii of stable states.
Energy possessed by the atom in the stable state is
e
2
W=
81t Eo r
Substituting the value ofr in Eq. (2.5), then
e
2
x 1tme
2
W = - ------:::--::--
81tE n
2
h
2
E
o 0
.......... ( 2.3 )
.......... ( 2.4 )
.......... ( 2.5 )
Junction Diode Characteristics 45
Thus energy W corresponds to only the coulombs force due to attraction between ground
electron (negative charge) and proton (positive charge).
Problem 2.1
Determine the radius of the lowest state of Ground State.
Solution
n = I
n
2
h
2
EO
r=
1tme
2
Plank's Constant, h = 6.626 x .} 0-
34
J-sec
Permitivity, Eo = 1O-
9
/361t
Substituting the values and simplifYing,
r = 0.58 A
O
2.1.4 ATOMIC ENERGY LEVELS
For different elements, the value of the free electron concentration will be different. By spectroscopic
analysis we can determine the energy level of an element at different wavelengths. This is the
characteristic of the given elem'ent.
The lowest energy state is called the normal level or ground level. Other stationary states
are called excited, radiating, critical or resonance levels.
Generally, the energy of different states is expressed in eV rather than in Joules, and the
emitted radiation is expressed by its wavelength A rather than by its frequency. This is only for
convenience since Joule is a larger unit and the energy is small and is in electron volts.
F= W2 - WI J
h
C
f= -
A
C = Velocity of Light = 3 x 10
10
cm / sec.
h = Plank's Constant = 6.626 x 10-
34
J - sec.
f= Frequency of Radiation
A = Wavelength of emitted radiation
W I and W
2
are the energy levels in Joules.
II eV = 1.6 x 10-
19
Joules I
C = (E2-EdxI.6xIO-19
A. h
(E2 - E
I
)>.< 1.6 x 10-
19
6.626 x 10-
34
46
Electronic Devices and Circuits
EI and E2 are energy levels in eY.
3x1010 x6.626x10-
34
or A = --------
(E2 - EI)x 1.6 x 10-
19
where A is in Armstrong, lAo = 10-
8
cm = 10-
10
m, and E2 and EI in eV.
2.1.5 PHOTON NATURE OF LIGHT
An electron can be in the excited state for a small period of 10-
7
to 10-
10
sec. Afterwards it
returns back to the original state. When such a transition occurs, the electrons will loose energy
equal to the difference of energy levels (E2 - E\). This loss of energy of the atom results in
radiation of light. The frequency of the emitted radiation is given by
f
= (E2 - E\)
h .
According to dassical theory it was believed that atoms continuously radiate energy. But
this is not true. Radiation of energy in the form of photons takes place only when the transition of
electrons will take place from higher energy state to lower energy state, so that ( E2 - EI ), is
positive. This will not occur if the transition is from lower energy state to higher energy state.
When such photon radiation takes place, the number of photons liberated is very large. This is
explained with a numercial example given below.
Problem 2.2
For a given 50 W energy vapour lamp. 0.1 % of the electric energy supplied to the lamp, appears
in the ultraviolet line 2,537 AO. Calculate the number of photons per second, of this wavelength
emitted by the lamp.
Solution
A = 12,400
(E2 - E\)
A = Wavelength of the emitted radiation.
E\ and E2 are the energy levels in eY.
(E2 - E\) is the energy passed by each photon in eV of wavelength A. In the given problem,
A = 2,537 AO. (E2 - E\) =?'
12,400
(E2 - E\) = 2,537 = 4.88 eV/photon
0.1 % of SOW energy supplied to the lamp is>,
0.1
I.e., 100 x 50 = 0.05 W = 0.05 J/Sec
1 W = 1 J/sec
Converting thi.., i ·to the electron Volts,
Junction Diode Characteristics
0.05 J /sec
1.6 x 10-
19
J/eV = 3.12 x 10
17
eV/sec
This is the total energy of all the photons liberated in the A = Wavelength of2,537 A 0.
Number of photons emitted per sec
total energy
energy / photon
3. 12 x 1 0
1
7 ev / sec
.
4.88ev / photon
= 6.4 x 10
16
photon/sec
The lamp emits 6.4 x 10
16
photons / sec of wavelength A = 2,537 A 0.
2.1.6 IONIZATION POTENTIAL
47
If the most loosely bound electron ( free electron) of an atom is given more and more energy, it
moves to stable state (since it is loosely bound, its tendency is to acquire a stable state. Electrons
orbiting closer to the nucleus have stable state, and electron orbiting in the outermost shells are
loosely bound to the nucleus). But the stable state acquired by the electron is away from the
nucleus ofthe atom. If the energy supplied to the loosely bound electron is enough large, to move
it away completely from the influence of the parent nucleus, it becomes detached from it.
The energy required to detach an electron is called Ionization Potential.
2.1.7 COLLISIONS OF ELECTRONS WITH ATOMS
If a loosely bound electron has to be liberated, energy has to be supplied to it. Consider the case
when an electron is accelerated and collides with an atom. If this electron is moving slowly with
less energy, and collides with an atom, it gets deflected, i.e., its direction changes. But no considerable
change occurs in energy. This is called Elastic Collision.
If the electron is having much energy, then this electron transfers its energy to the loosely
bound electron of the atom and may remove the electron from the atom itself. So another free
electron results. If the bombarding electron is having energy greater than that required to liberate
a loosely bound electron from atom, the excess energy will be shared by the bombarding and
liberated electrons.
Problem 2.3
Argon resonance radiation, corresponding to an energy of 11.6 e V falls upon sodium vapor. If a
photon ionizes an unexcited sodium atom, with what speed is the dectron ejected? The ionization
potential of sodium is 5.12 eY.
Solution
Ionization potential is the minimum potential required to liberate an electron from its parent atom.
Argon energy is 11.6 eY. Ionization Potential ofNa is 5.12 eY.
or Potential,
Its velocity
The energy possessed by the electron which is ejected is
11.6 - 5.12 = 6.48 eV
v = 6.48 volts ( .: 1 eV energy, potential is 1 V)
v = ~ ~ V = 5.93 x 10
5
.j6.48 = 1.51 x 10
6
m/sec.
48 Electronic Devices and Circuits
Problem 2.4
With what speed must an electron be travelling in a sodium vapor lamp in order to excite the yellow
line whose wavelength is 5,893 A 0.
Solution
12,400
Be> -- >2.11 eV
- 5,893 -
V = 2.11 Volts
( Corresponding to evergy of 2.11 e V,
the potential is 2.11 Volts)
Velocity, v = ~ ~ V = 5.93 x 10
5
.J2T1 = 8.61 x 10
5
m/sec.
Problem 2.5
A radio transmitter radiates 1000 Wat a frequency of 10 MHz.
Solution
(a) What is the energy of each radiated quantum in ev ?
(b) How many quanta are emitted per second?
(c) How many quanta are emitted in each period of oscillation of the
electromagnetic field?
(a) Energy of each radiated quantum = E = hf
f= 10 MHz = 10
7
Hz, h = 6.626 x 10-
34
Joules / sec
E = 6.626 x 10-
34
10
7
= 6.626 x 10-
27
Joules / Quantum
6.626 x 10-
27
-----::-- = 4.14 x 10-
8
eV / Quantum
1.6x10-
19
(b) 1 W = 1 Joule/sec
1000 W = 1000 Joules/sec = Total Energy
Energy possessed by each quantum = 6.626 x 10-27- Joules/Quantum.
1000 29
Total number of quanta per sec, N = 27 = 1.5 x 10 /sec
6.626x 10-
(c) One cycle = 10-
7
sec
Problem 2.6
Number of quanta emitted per cycle = 10-
7
x 1.51 x 10
29
= 1.51 x 10
22
per cycle
(a) What is the minimum speed with which an electron must be travelling in order
that a collision between it and an unexcited Neon atom may resuit in ionization
of this atom? The Ionization Potential of Neon is 21.5 V.
(b) What is the minimum frequency that a photon can have and still be a b l ~ to cause
Photo-Ionization of a Neon atom?
Junction Diode Characteristics
Solution
(a)
(b)
Ionization Potential is 21.5 V
Velocity = ~ 2 : V = 5.93 x 1 0
5
~ = 2.75 x 10
6
m/sec
Wavelength ofradiation,
12,400 12,400 0
A= = -- =577 A
E2 - EI 21.5
Frequency of radiation,
C 3x 10
8
j
- - = 5.2 x 10
15
Hz
- i - 577xlO-
IO
Prohlem 2.7
49
Show that the time for one revolution of the electron in the hydrogen atom in a circular path around
the nucleus is
Solution
But radius,
Prohlem 2.8
T = 4.J2 Eo (_ w)t .
v - ~ 4.:.'e
o
'
T = 2m = 2m(4mTI Eo r r
v e
e
2
2nrt(4nm Eo )'12
e
A photon of wavelength 1,400 A
0
is absorbed by cold mercury vapor and two other photons are
emitted. If one of these is the 1,850 A
0
line, what is the wavelength A of the second photon?
Solution
12,400 12,400
( E2 - EI ) = -1..-- = 1400 = 8.86 eV
1850° A
O
line is from 6.71 eV to 0 eY.
:. The second photon must be from 8.86 to 6.71 eY.
So, .1E = 2.15 e Y.
1..= 12,400
(E2 - E
I
)
12,400
A= -- = 5767Ao
2.15 .
50 Electronic Devices and Circuits
2.1.8 METASTABLE STATES
An atom may be elevated to an excited energy state by absorbing a photon of frequency 'f and
thereby move from the level of energy WI to the higher energy level W
2
where W
2
= WI + hf
But certain states may exist which can be excited by electron bombardment but not by photo
excitation (absorbing photons and raising to the excited state). Such levels are called metastable
states. A transition from a metastable level to a normal state with the emission of radiation has a
very low probability of occurence. Transition from higher level to a metastable state are permitted,
and several of these will occur.
An-electron can be in the metastable state for about 10-
2
to 10-
4
sec. This is the mean life
of a metastable state. Metastable state has a long lifetime because they cannot come to the
normal state by emitting a photon. Then if an atom is in metastable state how will it come to the
normal state? It cannot release a photon to come to normal state since this is forbidden. Therefore
an atom in the metastable state can come to normal state only by colliding with another molecule
and giving up its energy to the other molecule. Another possibility is the atom in the metastable
state may receive additional energy by some means and hence may be elevated to a higher energy
state from where a transition to normal state can occur.
2.1.9 WAVE PROPERTIES OF MATTER
An atom may absorb a photon of frequency f and move from the energy level W I to the higher
energy level W
2
where W
2
= WI + hf
Since a photon is absorbed by only one atom, the photon acts as if it were concentrated in
one point in space. So wave properties can not be attributed to such atoms and they behave like
particles.
Therefore according to 'deBroglie' hypothesis, dual character of wave and particle is not
limited to radiation alone, but is also exhibited by particles such as electrons, atoms, and molecules.
He calculated that a particle of mass 'm' travelling with a velocity v has a wavelength A given by
A = ~ = ~ ( A is the wavelength of waves consisting of these
mv p particles ).
where 'p' is the momentum of the particle. Wave properties of moving electrons can be made use
to explain Bohr's postulates. A stable orbit is one whose circumference is exactly equal to the
wavelength A or nA where n is an integer other than zero.
Thus 27t r = nA r = radius of orbit, n = Principle Quantum Number.
But according to DeBroglie,
h
A=-
mv
nh
27tr = -
mv
This equation is identical with Bohr's condition,
2.1.10 SCHRODINGER EQllATION
nh
mvr=-
27t
Schrodinger carried the implications of the wave nature of electrons. A branch of physics called
Wave Mechanics or Quantum Mechanics was developed by him. If deBroglie's concept of
Junction Diode Characteristics 51
wave nature of electrons is correct, then it should be possible to deduce the properties of an
electron system from a mathematical relationship called the Wave Equation or Schrodinger
Equation. It is
2
\7
2
<1> __ 1 . 8 <I> _
v2 8t
2
- 0, .......... (2.6.1)
2 8
2
8
2
8
2
where \7 =--+--+--
8x
2
8y2 8z
2
<I> can be a component of electric field or displacement or pressure. v is the velocity of the wave,
and 't' is the time.
The variable can be eliminated in the equation by assuming a solution.
<I> (x, y, z, t) = \jJ (x, y, z)e
jwt
This represents the position of a particle at 't' in 3-D Motion.
co = Angular Frequency = 2 1tf
But
co is regarded as constant, while differentiating.
<I> is a function of x, y, z and t.
\jJ is a function of x, y and z only.
To get the independent Schrodinger Equation
8<1>
at = jro\jJ(x, y, z)eJwt
But
So,
8
2
<1>
8t
2
= - ro
2
\jJ(x,y,z)e.lCl)t
co = 27Tf
co
2
= 41t2j
Substituting th is in the original Schrodinger's Equation,
e
iwt
{\7
2
\jJ + v
12
x 41t2f2\jJ} = °
v Velocity
But ')... = -. Wavelength =
f ' Frequency
But
2 41t2
\7 \jJ+-\jJ = °
')...2
h h
')...=-=-
mv p
p=mv
This is deBrogile's Relationship.
p2
')...2 = h2
v = Velocity.
.......... (2.6.2)
52 Electronic Devices and Circuits
But p2 = m
2
v
2
= 2 ( Kinetic Energy )m .: K.E. = Y2mv2
Kinetic Energy = Total Energy (W) - Potential Energy (U)
p2 = 2 ( W - U ) . m
p2 2m
h2 = h2 ( W - U ) .......... (2.6.3)
Substituting Eq. (2.7.3) in Eq. (2.7.2) we get,
87t
2
m
V
2
\{' + -2- (W - U) \{' = 0.
h
This is the Time Independent Schrodinger Equation \{' is a function of't' But this equation as
such is not containing the term 't'.
2.1.11 WAVE FliNCTION
\{' is called as the wave function, which describes the behavior of the particle. \{' is a quantity
whose square gives the probability offinding an electron. I\{' 12( dx x dy x dz) is proportional to
the probability of finding an electron in volume dx, dy and dz at point P(x, y, z).
Four quantum numbers are required to define the wave function. They are:
1. Tile Principal Quantum Number 'n' :
It is an integer 1,2,3, .... This number determines the total energy associated
with a state. It is same as the quantum number 'n' of Bohr atom.
2. Tile Orbital Angular Momentum Quantum Number I:
It takes values 0, I, 2 ... 1. .. (n -1)
The magnitude of this angular momentum is .J(iXI + 1) x J:..
27t
It indicates the shape of the classical orbit.
3. Tile orbital magnetic number m,:
This will have values 0, ± I, ±2 .... ± I. This number gives the orientation of
the classical orbit with respect to an applied magnetic field.
The magnitude of the Angular Momentum along the direction of magnetic
field = m r ( ; ~ ) .
4. Electron Spin:
It was found is 1925 that in addition to assuming that electron orbits round
the nucleus, it is also necessary to assume that electron also spins around itself,
in addition to orbiting round the nucleus. This intrinsic electronic angular
momentum is called Electron Spin.
When an electron system is subjected to a magnetic field, the spin axis will
orbit itself either parallel or anti-parallel to the direction of the field. The electron
angular momentum is given by ms (2:) where the spin quantum ms number
may have values +Y2 or -Y2.
Junctil}n Diode Characteristics
2.1.12 ELECTRONIC CONFIGURATION
PAlIU'S EXCUISION PRINCIPLE
53
No two electrons in an electron system can have the same set of four quantum numbers n, I, mt
and ms.
Electrons ".vill occupy the lower most quantum state.
ELECTRONIC SHELLS ( PRINCIPLE QUANTLIM NllMBER )
All the electrons which have the same value of on' in an atom are said to belong to the same
electron shell. These shells are identified by letters K, L, M, N corresponding to n = 1,2,3,4 .....
A shell is subdivided into sub shells corresponding to values of 1 and identified as s, p, d, f, g; h,. for
1 = 0, I, 2, 3 ... respectively. This is shown in Table 2.1.
Table 2.1
Shell ..... K L M
n ..... 1 2 3
1 ..... 0 0 I 1 0 1
subshell .....
I
s S
I
P
s
P
No. of 2 2
I
6 2
I
6
I
electron 2 8 18
ELECTRON SHELLS AND SllBSHELLS
Number of Electrons in a sub shell = 2(21 + I)
n = I corresponds to K shell
1 = ° corresponds to s sub shell
1 = 0, ... , (n - I) if n = I,
1= ° is the only possibility
N
4
2 0 1 2
d s
P
d
10 2
I
6
I 10 J
32
Number of electrons in K shell = 2(21 + I) = 2(0 + I) = 2 electrons.
K shell will have 2 electrons.
3
f
14
This is written as I s2 pronounced as "one s two" ( I corresponds to K shell, n = I; s is the
sub shell corresponds to 1 = ° number of electron is 2. Therefore, I s2 )
If n = 2, it is 'I' shell
If 1 = 0, it is's' sub shell
If 1 = I, it is 'p' sub shell
Number of electron in's' sub shell ( i.e., 1 = 0) = 2 x (21 + I) = 2(0 + I) = 2
Number of electron is 'p' sub shell (i.e., 1 = I) = 2[ (I x 2) + I)] = 6
In 1 shell there are two sub shells, sand p.
Total number of electron in 1 shell = 2 + 6 = 8
This can be represented as 2s2 2p6
54 Electronic Devices and Circuits
If n = 3, it is M shell
It has 3 sub shell s. p, d, corresponding to 1 = 0, 1, 2
In's' sub shell number of electrons (l =0) = 2( :. 1= 0) 2(21 + 1)
In 'p' sub shell number of electrons (/ = 1) = 2(2 + 1) = 6
In 'd' sub shell number of electrons (l =2) = 2(2 x 2 + 1)= 10
. . Total number of electrons in M shell = 10+ 6 + 2 = 18
This can be represented as 3s
2
3p6 3d
1o
Is2 2s2 2p6 3s
2
3p6 3d
1o
4s
2
4p6 4d
lO
ELECTRONIC CONFIGllRATION
Atomic number 'z' gives the number of electrons orbitin§ round the nucleus. So from the above
analysis, electron configuration can be given as 1 s2 2s2 2p 3s I. First k shell (n = 1 ) is to be filled.
Then I shell ( n = 2 ) and so on.
In 'k' shell there is one sub shell (s) (/= 0). This has to be filled.
In 'L' shell there are two sub shells and p. First - s and then p are to be filled and so on.
The sum of subscripts 2 + 2 + 6 + I = 11. It is the atomic number represented as Z.
For Carbon, the Atomic Number is 6., i.e., Z = 6.
. . The electron configuration is I s2 2s2 2p2
For the Ge Z = 32. So the electronic configuration of Germanium is,
.. Is2 2s2 2p6 3s
2
3p6 3d
1o
4s
2
4p2
For the Si Z = 14. So the electronic configuration of Silicon is,
Is2 2s2 2p6 3s
2
3p2
2.1.13 TVPES OF ELECTRON EMISSION
Electrons at absolute zero possess energy ranging from 0 to EF the fermi level. It is the characteristic
of the substance. But this energy is not sufficient for electrons to escape from the surface. They
must posses energy EB = EF + Ew where Ew is the work function in e V.
E
B
Barrier's Energy
EF = Fermi Level
Ew = Work Function.
Different types of Emission by which electrons can emit are
(1) Thermionic Emission
(2) Secondary Emission
(3) Photoelectric Emission
(4) High field Emission.
1. THERMIONIC EMISSION
Suppose, the metal is in the form of a filament and is heated by passing a current through it. As the
temperature is increased, the electron energy distribution starts in the metal changes. Some electrons
may acquire energy greater than EB sufficient to escape from the metal.
Junction Diode Characteristics 55
Ew : Work function of a metal. It represents the amount of energy that must be given
for the electron to be able to escape from the metal.
It is possible to calculate the number of electrons striking the surface of the metal per
second with sufficient energy to be able to surmount the surface barriers and hence escape.
Based upon that, the thermionic current is,
Ith = S x Ao T2 e-
E
" IKT
It is alsb written as
where
where
I ~ - 8
...!h. = J = AoT2 e kT = AT2 e T
S
Ew
A=Ao and B= k
S = Area of the filament in m
2
( Surface Area)
Ao= Constant whose dimensions are A/m
2
oK
T = Temperature in oK
K = Boltzman's constant eV/oK
Ew = Work function in e V
.......... ( 2.7 )
This equation is called Thermionic Emission Current or Richardson - Dushman Equation.
Ew is also called as latent heat of evaporation of electrons similar to evaporation of molecules
from a liquid.
Taking logarithms
Ew 2
log Ith = log (S A
o
) - kT log e + log T
log Ith - 2 log T = log S Ao - 0.434 ( ~ ; )
log e = 0.434
So if a graph is plotted between (log Ith - 2 log T) V s ~ . the result will be a straight line
T
having a slope = - 0.434 ( ~ ; ) from which Ew can be determined.
.. Ith and T can be determined experimentally.
Ith is a very strong function ofT. For Tungsten. Ew = 4.52 eY.
CONTACT POTENTIAL
Consider two metals in contact with each other formingjunction at C as in Fig 2.1 . The contact
difference of potential is defined as the Potential Difference V AB between a point A, just out side
metal 1 and a point B just outside metal2. The reason for the difference of potential is. when two
metals are joined, electrons will flow from the metal of lower Work Function. say 1 to the metal of
higher Work Function say 2. ( .,' Ew = EB - EF electrons of lower work function means Ew is
small or EF is large). Flow of electrons from metal I to 2 will continue till metal 2 has acquired
56 Electronic Devices and Circuits
sufficient negative charge to repel extra new electrons. EB value will be almost same for all
metals. But EF differs significantly.
A B
2
c
Fig 2.1 Contact Potential
In order that the fermi levels of both the metals are at the same level, the potential energy
difference EAB = EW2 - Ew], that is, the contact difference of potential energy between two
metals is equal to the difference between their work functions.
Ifmetals I and 2 are similar then contact potential is zero. If they are dissimilar, the metal
with lower work function becomes positive, since it looses electrons.
ENERGIES OF EMITTED ELECTRONS
At absolute zero, the electrons will have energies ranging from zero to E
F
. So the electrons
liberated from the metal surface will also have different energies. The minimum energy required
is the barrier potential to escape from the metal surface, But the electrons can acquire energy
greater than EB to escape from the surface. This depends upon the initial energy, the electrons are
possessing at room temperature.
Consider a case where anode and cathode are plane parallel. Suppose the voltage applied to
the cathode is lower, and the anode is indirectly heated. Suppose the minimum energy required to
escape from the metal surface is 2 eY. As collector is at a potential less than 2V, electrons will be
collected by the collector. If the voltage of cathode is lower, below 2V, then the current also
decreases exponentially, but not abruptly. If electrons are being emitted from cathode with 2eV
energy, and the voltage is reduced below 2V, then there must be abrupt drop of current to zero.
But it doesn't happen This shows that electrons are emitted from surface of the emitter with
different velocities. The decrease of current is given by I = hh.e-Vt where V
t
is the retarding
potential applied to the collector and V
T
is Volt equivalent of temperature.
kT T
VT = --;- = 11,600 .......... (2.8)
T = Temperature in OK
K = Boltzman's Constant in JfK
SCHOTTKY EFFECT
If a cathode is heated, and anode is given a positive potential, then there will be electron emission
due to thermionic emission. There is accelerating field, since anode is at a positive potential. This
accelerating field tends to lower the Work Function of the cathode material. It can be shown that
under the condition of accelerating field E is
I = Itlt e +0.44 &l/2/T
where Ith is the zero field thermionic current and T is cathode temperature in 0 OK.
The effect that thermionic current continues to increase as E is increased ( even though T
is kept constant) is known as Schottky Effect.
Junction Diode Characteristics 57
2. SECONDARY EMISSION
This emission results from a material (metal or dielectric) when subjected to electron bombardment.
It depends upon,
I. The energy of the primary electrons.
2. The angle of incidence.
3. The type of material.
4. The physical condition of surface; whether surface is smooth or rough.
Yield or secondary emission ratio S is defined as the ratio of the number of secondary
electrons to primary electrons. It is small for pure metals, the value being 1.5 to 2. By contamination
or giving a coating of alkali metal on the surface, it can be improved to 10 or 15.
3. PHOTO ELECTRIC EMISSION
Photo-Electric Emission consists of liberation of electrons by the incidence of light, on certain
surfaces. The energy possessed by photons is hfwhere h is Plank's Constant andfis frequency
of incident light. When such a photon impinges upon the metal surface, this energy lif gets
transferred to the electrons close to the metal surface whose energy i ~ , very near to the barrier
potential. Such electrons gain energy, to overcome the barrier potential and escape from the surface
of the metal resulting in photo electric emission.
For photoelectric emission to take places, the energy of the photon must at least be equal to
the work function of the metal. That is hf ~ e$ where $ is the voltage equivalent Work Function,
(i.e. Work Function expressed in Volts). The minimumfrequency that can cause photo-electric
emission is called threshold frequency and is given by
e<l>
/r=-
h
The wavelength corresponding to threshold frequency is called the Threshold Wavelength.
A = ~ = ~
t f
t
<l>e
If the frequency ot"radiation is less than/r. (hen additional energy appears as kinetic energy
of the emitted electron
hf= $e + ~ m v
2
$ is the Volt equivalent of Work Function.
v = ~ 2 : V
v is the Velocity of electrons in m/sec.
1111= = $e + eV 1
.......... ( 2.9 )
LAWS OF PHOTO ELECTRIC EMISSION
1. For each photo sensitive material, there is a threshold frequency below
which emission does not take place.
2. The amount of photo electric emission (current) is proportional to
intensity.
58 Electronic Devices and Circuits
3. Photo electric emission is instantaneous. (But the time lag is in
nano-sec ).
4. Photo electric current in amps/watt of incident light depends upon 1'.
4. HIGH FIELD EMISSION
Suppose a cathode is placed inside a very intense electric field, then the Potential Energy is
reduced. For fields of the order of 10
9
V 1m, the barrier may be as thin as 100 A 0. So the electron
will travel through the barrier. This emission is called as High Field Emission or Auto Electronic
Emission.
Problem 2.9
Estimate the percentage increase in emission from a tungsten filament when its temperature is
raised from 2400 to 2410 oK.
Ao = 60.2 x 10
4
Alm
2
/°K2
B = 52.400 oK
A and B are constants tn the equation for current density J
Solution
JS
I
=AT
I
2
= 1142A/m2
JS
2
= ATl e -8/T = 1261 A 1m
2
.
1261-1142 xl 00 - 0
Percentage Increase = 1142 - 10.35Yo
Problem 2.10
A photoelectric cell has a cesium cathode. When the cathode is illuminated with light of A. = 5500
x 10-
10
m, the minimum anode voltage required to inhibit built anode current is 0.55 V. Calculate
( a) The work function of cesium
( b) The longest A. for which photo cell can function.
by applying -0.55V to anode the emitted electrons are repelled. So the current can be inhibited
Solution
(a) 11/= e<l> + eV V = 0.55 volts <I> = Work Function (WF) = ?
C
f= -
A.
h.C
- =e(<I>+V)
A.
Plank's Constant, h = 6.63 x 10-
34
J sec.
Charge of Electron,e = 1.6 x 10-
19
C
Velocity of Light, C = 3 x 10
8
m/sec
6.63 x 1 0-
34
x 3 x 10
8
5500x10-
10
<I> = 1.71 Volts
= 1.6 x 10-
19
(<I> + 0.55)
Junction Diode Characteristics
(b) Threshold Wavelength :
Problem 2.11
Ch
Ao =
12,400 12 400
A
o
= -<1>- = -f:71 = 7,250 A
O
59
If the temperature of a tungsten filament is raised from 2300 to 2320 OK, by what percentage will
the emission change? To what temperature must the filament be raised in order to double its
energy at 2300 OK. Ew for Tungsten = 4.52 eY. Boltzman's Constant K = 8.62 x 10
5
eV 1
0
K.
Solution
(a) Ith = S.Ao T2 e-Ew/kT
Taking Logarithms,
Differentiating,
E",
In Ith - 2 log T = In S Ao - kT
(
2 Ew)dT
T + KT2 T2
di
th
_ (2 + Ew) dT _ (2 + 4.52 ) _ 0
Ith - kT T - 2310 - 21.4%
--452
--452
21th = S Ao (T)2 e8.62xlO-5
Ratio of these two equation is
(
T )2 _ 52.400 +22 8
2 = 2300 e T
Taking log to the base 10,
log 2 = 2 log - (52;00 + 22.8 )lOge
(
T} 52400
log 2 = 2 log 2300 - -T- x 0.434 - 22.8 x 0.434
(
T) 22,800
9.6 + 2 log 2300 = -T-
This is solved by Trail and Error Method to get T = 2370
0
K
60 Electronic Devices and Circuits
Problem 2.12
In a cyclotron, the magnetic field applied is 1 Tesla. If the ions (electrons) cross the gap between
the 0 shaped discs dees twice in each cycle, determine the frequency of the R.F. voltage. If in
each passage through the gap, the potential is increased by 40,000 volts how many passages are
required to produce a 2 million volts particle? What is the diameter of the last semicircle?
Solution
B = 1 Tesla = 1 Wb/m
2
35.51-l sec
Time taken by the particle to describe one circle is T = B
35.5 X 10-
6
T = 1 sec,
when it describes one circle, the particle passes through the gap twice.
The frequency of the R.F voltage should be the same.
j
- 1 _ 1 = 2.82 x 10
10
Hz
- T - 35.5 X 10-6
Number of passages required:
In each passage it gains 40,000 volts.
2x 106 y
To gain 2 million electron volts = 1 = 50
40x 10- Y
Diameter of last semicircle:
The initial velocity for the 50
th
revolution is the velocity gained after 49
th
revolution.
Accelerating potential V
o
= 49 x 40,000 = 1,960 KV 49
th
revolution
. . . 3.37x 10-
6
~
RadIUs of the last semIcIrcle = R = "" Yo
B
R= 3.37xlO-
6
~ 1 9 6 0 x 1 0 3 =4718 x 10-4
m
1 .
Diameter = 2R = 94.36 x 10-4 m
Problem 2.13
The radiated power density necessary to maintain an oxide coated filament at 1 lOoK is found to
be 5.8 x 10
4
W/m
2
. Assume that the heat loss due to conductor is 10% of the radiation loss.
Calculate the total emission current and the cathode efficiency 11 in ma/w.
Take
Solution
Ew = 1.0 eV,
A
o
= 100 A/m
2
I °K2
S = 1.8 cm
2
K = Boltzman's Constant in eV I oK = 8.62 x 10- 5eV I oK
Power Density = 5.8 x 10
4
W/m
2
Junction Diode Characteristics 61
If 10% is lost by conductor, then the total input power density is 1.1 x 5.8 x 10
4
= 6.3 8 W /
m
2
and the total input power is 6.38 x 10
4
x Area (1.8 x 10--4) = 11.5 W.
Ith = SAo T2 e-Ew/kT
= l.8 x 10--4 x 100 x (1100)2 e 862xlO-
s
xllOO
Ith = 2.18 x 10
4
e-
10
.
55
Ith = 0.575 A
0.575
Cathode Efficiency, 11 = l1.5 = 50 rnA / oK
2.2 ENERGY BAND STRUCTURES
E
r
Insulator is a very poor conductor of electricity. Ex: Diamond
Resistivity, p > 10
9
n - cm
Free electron Concentration, n:: 10
7
electrons/m
3
Energy Gap, EG» 1 e V
Metal is an excellent conductor of electricity. Ex: AI, Ag, Copper
p < 10-
3
n - cm
Free electron concentration, n = 10
28
electrons/m
3
Energy Gap, EG = 0
Semiconductor is a substance whose conductivity lies between these two. Ex: C, Si, Ge,
GaAs etc.,
E E
Conduction Band
(C.B)
i
Conduction Band
(C. B)
i
Forbidden Band
:>
Q)
(F.B)
Valence Band
Valence Band
(Y.B)
(Y. B)
~ (Wave Vector) ~ K ~ K ~ K
Metal (Conductor) C.B
and V.B overlap
Insulator Semiconductor
Fig 2.2 Energy band structures.
62 Electronic Devices and Circuits
2.2.1 INSULATOR
For diamond, the value of energy gap Eg is 6eY. This large forbidden band separates the filled
valence band region from the vacant conduction band. The energy, which can be supplied to an
electron from an applied field, is too small to carry the particle from the filled valence band into the
vacant conduction band. Since the electron can not acquire externally applied energy, conduction
is impossible and hence diamond is an insulator.
2.2.2 SEMICONDUCTOR
For Semiconducting materials, the value of energy gap Eg will be about 1 e Y.
Germanium ( Ge ) has Eo = 0.785 eV, and Silicon is 1.21 eV at OOK. Electron can not
acquire this much Energy to travel from valence band to conduction band. Hence conduction will
not take place. But Eo is a function of temperature T. As T increases, Eo decreases.
For Silicon ( Si ) &0 decreases at the rate of 3.6 x 10-4 eV fK
For Germanium Eo decreases at the rate of 2.23 x 10-4 eVfK
For Si, Eo = 1.21- 3.6 x 10-4 x T
For Ge, Eo = 0.785 - 2.23 x 10-4 x T.
The absence of an electron in the semiconductor is represented as hole.
2.2.3 METAL (CONDUCTOR)
In a metal, the valence band may extend into the Conduction Band itself. There is no forbidden
band, under the influence of an applied field, the electron will acquire additional energy and move
into higher states. Since these mobile electrons constitute a current, this substance is a conductor.
When an electron moves from valence band into conduction band in a metal, the vacancy so
created in the valence band can not act as a hole. Since, in the case of metals, the valence
electrons are loosely bound to parent atom. When they are also in conduction, the atom can pull
another electron to fill its place.
In the energy band diagram, the y-axis is energy. The x-axis is wave vector K, since the
energy levels of different electrons are being compared. The Ge has structure of,
Is2 2s2 2p6 3s
2
3
p
6 3d
lO
4s
2
4p2
The two electrons in the s sub shell and 2 in the p sub shell are the 4 electrons in the
outermost 4th shell. The Germanium has a crystalline structure such that these 4 electrons are
shared by 4 other Germanium atoms. For insulators the valence shell is completely filled. So there
are no f r ~ e electrons available in the outermost shell. For conductors say Copper, there is one
electron 10 the outermost shell which is loosely bound to the parent atom. Hence the conductivity
of Copper is high.
2.3 CONDUCTION IN SEMICONDUCTORS
Conductors will have Resistivity
Insulators will have Resistivity
p < 10-
3
O-cm;
P 2: 10
9
O-cm;
n = 10
28
electrons/m
3
n = 10 7 electrons/m
3
For semiconductors value lies between these two.
Some of The different types of solid state devices are :
1. Semiconductor Diode Zener diode, Tunnel diode, Light emitting
diode; Varactor Diode
2. SCR Silicon Controlled Rectifier
Junction Diode Characteristics
3. Transistor ( BJT)
4. FET
5. MOSFET
5. UJT
6. Pltoto Transistors
7. IMPATT
8. TRAPATT
9. BARRITT
Advantages 0/ Semiconductor Devices
1. Smaller in size.
PNp, NPN
Field Effect Transistors
Metal Oxide Semiconductor Field Effect
Transistors
Unijunction Transistor
Impact Ionisation Avalanche Transit Time
Device
Trapped Plasma Avalanche Transit Time
Device
Barrier Injected Avalanche Transit Time
Device
2. Requires no cathode heating power (warm up time compared to Vacuum
Tubes ).
3. They operate on low DC power.
4. They have long life. (Tubes will pop up frequently).
Disadvantages
1. Frequency range of operation is low.
2. Smaller power output.
3. Low permissible ambient temperature.
4. Noise is more ( because of recombination between holes and electrons ).
63
•
In a metal, the outer or valence electrons of an atom are as much associated with one ion
(or parent atom) as with another, so that the electron attachment to any individual atom is almost
zero. In other words, band occupied by the valence electrons may not be completely filled and that
these are forbidden levels at higher energies. Depending upon the metals at least one and sometimes
two or three electrons per atom are free to move throughout the interior of the metal under the
action of applied fields.
Germanium semiconductor has four valence electrons. Each of the valence electrons of a
germanium atom is shared by one of its four nearest neighbours. So covalent bonds result. The
valence electrons serve to bind one atom to the next also result in the valence electron being tightly
bound to the nucleus. Hence in spite of the availability of four valence electrons, the crystal has
low conductivity.
In metals the binding force is not strong. So free electrons are easily available.
According to the electron gas theory of a metal, the electrons are in continuous motion. The
direction of flight being changed at each collision with the impinging (almost stationary) ions. Tlte
average distance between collisions is called tlte mean/ree path. Since the motion is random,
on an average there will be as many electrons passing through unit area in the metal in any
directions as in the opposite directions, in a given time. Hence the average current is zero, when
no electric field is applied.
64 Electronic Devices and Circuits
Now, if a constant electric field 'I>' V /m is applied to the metal, as a result of the electrostatic
force, the electrons would be accelerated and the velocity would increase indefinitely with time if
the electron will not collides with any other particle. But actually the electron collides with number
of ions and it loses energy. Because of these collisions, when the electron loses its energy, its
velocity will also decrease. So finally a steady state condition is reached where an infinite value of
drift velocity 'v' is attained. This drift velocity is in the direction opposite to that of the electric field
and its magnitude is proportional to the electric field E. Thus velocity is proportional to E or v = IlE,
where Il is called the mobility of electrons.
v ex: I> or v = IlE
V is in m/sec. E is in vim.
v mxm 2
Il::; - = = m / volts-sec
E Secx Volts
Mobility is defined as the velocity per unit electric field. The units are m
2
lv-sec.
Because of the thermal energy, when no electric field is applied, the motion of electrons is in
random nature. When an electric field is applied, steady state drift velocity is superimposed upon
the random thermal motion of the electrons. Such a directed flow of electrons constitute a current.
If the concentration of free electrons is On' (electrons/m
3
) the current density is J, (amps/m2)
J = ne v .......... ( 2.10 )
But velocity
IV=/lE
J = ne /lE
But ne Il = cr (sigma)
cr is the conductivity of the metal in (O-mr' or mhos/m.
So IJ=crxEI .......... ( 2.11 )
As temperature increases mobility decreases. This is analogous to large number of people
in a small room. The movement of each person is restricted. If less number of persons are there
the movement is large or mobility is large. The electrons acquire from the applied field, as a result
of collisions is given to the lattice ions. Hence power is dissipated in the metal.
From the above expression for conductivity, cr is proportional
to on', the number of free electrons. For a good conductor 'n'
is very laTlle, -lri
8
electronslm
3
. For an insulator n is very
small-10
7
electronslm
3
. For a semiconductor the
value of 'n' lies between these two. The valance electrons in a
semiconductor are not free to move like in a metal, but they are
trapped between two adjacent ions.
Germanium has a total of 32 electrons in its atomic
Electron
\ Nucleus
0/ 0
0
·' - - - - - - - - - - - - - - - - ~ - - ; O · 0
Q - ~ t - - - - - - - - - - - - - ~ - - " 0
AI A,
Fig 2.3 Electronic structure.
structure, arranged in shells. Each atom in a Germanium crystal contribute four valence electrons.
The boundry forces between neighbours attains result from, the fact is that each of the valence
electrons of a germanium atom is shared by one of its four nearest neighbours.
Electron e, of atom A, is bound by atom A
2
. Electron e2 of atom A2 is bond by the atom
AI. So because of this, covalent bonds result. The fact that valence electrons serve to bound
atom to the next also results in the valence electrons being bound to the nucleus. Hence inspite of
having four valence electrons, the crystal has low conductivity ( Fig. 2.3 ).
Junction Diode Characteristics 65
At very low temperatures say 0° K, the ideal structure is achieved and the semiconductor
behaves as an insulator, since no free carriers of electricity are available. However at room
temperature, some of the covalent bonds will be broken because of the thermal energy supplied to
the crystal, and conduction is made possible. An electron which for the greater period of time
forms part of a covalent bond, is shown as being dislodged and so free to wander in a random
fashion throughout the crystal.
1,2,3, and 4, are Ge atoms with 4 valance electrons (Fig.
o 0@ 2.4). These four valence electrons of each atom are shared by
: • '. other atoms and so bounded by covalent bonds, except for atom
; :)'01' I. The electron of Ge atom I is dislodged from its original
: 0.( Ifr .. Electron position, because of the thermal energy and so the covalent bonds
o • _ ••• '. • ..:..... 0 bonding the electron are broken. So this electron is now a free
o (J.: .... :. 0'. : ... _ : _ 0 0 electron to wander anywhere in the material ti II it collides with
o @ : • :Q) ® 0 some other atom. The absence of the electron in the covalent
..
: • bond is represented by a hole. This is the concept of hole. The
: • : energy required to break different covalent bonds will be different.
. '
~ So at a time at room temperature all the covalent bonds are not
o \:..)0 ®
broken to create innumerable free electrons.
Fig 2.4 Covalent bonds The energy Eg required to break such a covalent bond is
about 0.72 eV for Germanium and 1.1 eV for silicon (at room
temperature). A hole can serve as a carrier of electricity. Its significance lies in this characteristics.
The explanation is given below:
When a bond is incomplete so that hole exists, it is relatively easy for a valence electron in
a neighbouring atom to leave its covalent bond to fill this hole. An electron moving from a bond to
fill a hole leaves a hole in its initial position. Hence the hole effectively moves in the direction
opposite to that of the electron. Thus hole in its new position may now be filled by an electron
from another covalent bond and the hole will correspondingly move one more step in the direction
opposite to the motion ofthe electron. Here we have a mechanism for the conduction of electricity
which does not involve free electrons. Only the electrons are exchanging their position and there
by current is flowing. In other words, the current is due to the holes moving in the opposite
direction to that of electrons. To explain this further,
2 3 4 5 6 7 8 9 10
(a) 0 0 0 0 0 0 0 0 0 0
0000000000
(b) 2 3 4 5 6 7 8 9 10.
Fig. 2.5 Current flow by the movement of holes
In row (a) there are 10 ions in Fig. 2.5. Except for 6, all the covalent bonds of the ions are
intact. The ion 6 has a broken covalent bond or one of its valence electrons got dislodged. So the
empty place denotes a hole. Now imagine that an electron from ion 7 moves into the hole at ion
6. Then the configuration is as shown in row (b). Ion 6 in the row is completely filled. There is no
-broken covalent bond. But ion 7 has a vacancy now, since it has lost one of its valence electrons.
66 Electronic Devices and Circuits
Effectively the hole has moved from ion 6 to ion 7. So the movement of holes is opposite to that of
electrons. The hole behaves like a positive charge equal in magnitude to the electron charge.
In a pure semiconductor the number of holes is equal to the number of free electrons.
Thermal agitation continues to produce new electron hole pairs whereas some other hole electron
pairs disappear as a result of recombination.
This is analogous to passengers travelling in a bus. Bus is the semiconducting material.
Standing passenger are free electron. When a sitting passenger gets down, a hole is created. This
hole is filled by a free electron that is a standing passenger. This process goes on as the bus is
moving from stage to stage. If there are many standing passengers, without any vacant seat, it is
analogous to n-type semiconductor. Ifthere are many seats vacant without any standing passenger
it is like a p-type semiconductor.
So the semiconductros are classified as :
n-type Sefiliconductor
p-type Semiconductor
Free electron concentration 'n' is greater than
hole concentration. D > p.
Hole concentration 'p' is greater than free electron
concentration. p > D.
Instrinsic Semiconductor D = p.
2.4 CONDUCTIVITY OF AN INTRINSIC SEMICONDUCTOR
When valence electrons are exchanging theIr posItions, we say holes are moving. Current is
contributed by these holes current is nothing but rate of flow of charge. Holes are positively
charged. So hole movement contributes for flow of current. Because of the positive charge
movement, the direction of hole current is same as that of conventional current. Suppose to start
with, there are many free electrons, and these wiIl be moving in random directions. A current is
constituted by these electrons. So at any instant, the total current density is summation of the
current densities due to holes and electrons. The charge of free electrons is negative and its
mobility is J..ln. The hole is positive, and its mobility is J..lp. The charge of both holes and electrons
are same 'e'. A hole can move from one ion to the nearest where as an electron is free to move
anywhere till it collides with another ion or free electron. Electrons and holes move in opposite
directions in an electric field E. Though they are of opposite sign, the current due to each ion is in
the same direction.
Current Density, J = (JE
J = (n J..ln + pJ..lp) x ex E = (J x E
n = Magnitude of Free Electron Concentration
p = Magnitude of Hole Concentration
(J = Conductivity n / cm or Seimens
(J = (n J..ln + P J..lp) e = neJ..ln + peJ..lp
A pure or intrinsic semiconductor is one in which n = p = nj where n
l
is intrinsic concentration.
In a pure Germanium at room temperature, there is about one hole-electron pair for every
2 x 10
9
Germanium atoms. As the temperature increases, covalent bonds are broken and so more
free electrons and holes are created. So n
l
increases, as the temperature increases, in accordance
with the relationship,
Junction Diode Characteristics
3
n.'= AxT2 e-EGo/2kT
I
EG = Energy Gap in ev.
e = Charge of an Electron or hole
A = Constant for a semiconductor;
for Ge A = 9.64 x 10
21
;
for Ge EG = 0.785 eV at 0 oK
forSi EcJ= 1.21 eV at 0 oK
EG at room temperature,
for Ge = 0.72 eV
for S i = 1.1 e V
T = Temperature in 0 oK
k = Boltzman 's Constant.
67
.......... (2.12)
As n
l
increases with T, the conductivity also increases, with temperature for semiconductor.
In other words, the resistivity decreases with temperature for semiconductor. On the other
hand resistance increases with temperature for metals. This is because, an increase in
temperature for metals results in greater thermal motion of ions and hence decrease in the mean
free path of the free electrons. This results in decrease of the mobility of free electrons and so
decrease in conductivity or increase in resistivity for metals.
2.5 DONOR TYPE OR n-TYPE SEMICONDUCTORS
Intrinsic or pure semiconductor is of no use since its conductivity is less and it can not be charged
much. If a pUie semiconductor is doped with impurity it becomes extrinsic. Depending upon
impurity doped, the semiconductor may become n-type, where electrons are the majority carriers
or donor type, since it donates an electron. On the other hand if the majority carriers are holes,
it is p-type or acceptor type semiconductor, because it accepts an electron to complete the
broken covalent bond.
Germanium atom with its electrons arranged in shells will have configuration as
Is2 2s2 2p2 3s
2
3p6 3d
1o
4s
2
4p2
Ge is tetravalent (4). 'Ge' becomes n-type if a pentavalent (5), impurity atoms such as
Phosphorus (P), or Arsenic are added to it.
The impurity atoms have size of the same order as that ofGe atoms. Because of the energy
supplied while doping, the impurity atom dislodges one from its normal position in the crystal
lattices takes up that position. But since the concentration of impurity atoms is very small (about
I atom per million ofGe atoms), the impurity atom is surrounded by Ge atoms. The impurity atom
is pentavalent. That is, it has 5 electrons in the outermost orbit (5 valence electrons). Now 4 of
these are shared by Ge atoms, surrounding the impurity atom and they form covalent bonds. So
one electron of the impurity atom is left free. The energy required to dislodge this fifth electron
from its parent impurity atom is very little of the order of 0.01 eV to 0.05 eY. This free electron is
in excess to the free electrons that will be generated because of breaking of covalent bonds due to
thermal agitation. Since an excess electron is available for each impurity atom, or it can denote
an electron it is called n-type, or donor type semiconductor.
•
\ •. J
68 Electronic Devices and Circuits
2.6 ACCEPTOR TYPE OR P - TYPE SEMICONDUCTORS
An intrinsic semiconductor when doped with trivalent (3) impurity atoms like Boron, Gallium Indium,
Aluminium etc., becomes p-type or acceptor type.
Because of the energy supplied while doping, the impurity atom dislodges ane Ge atom from
the crystal lattice. The doping level is low, i.e., there is one impurity atom for one million Ge atoms,
the impurity atom is surrounded by Ge atom. Now the three valence electrons of impurity atom
are shared by 3 atoms. The fourth Ge atom has no electron to share with the impurity atom. So
the covalent bond is not filled or a hole exists. The impurity atom tries to steal one electron from
the neighboring Ge atoms and it does so when sufficient energy is supplied to it. So hole moves.
There will be a natural tendency in the crystal to form 4 covalent bonds. The impurity atom (and
not just 3) since all the other Ge atoms have 4 covalent bonds and the structure ofGe semiconductor
is crystalline and symmetrical. The energy required for the impurity atom to steal one Ge electron
is 0.0 I e V to 0.08 e Y. This hole is in excess to the hole created by thermal agitation.
2.7 IONIZATION ENERGY
Ifintrinsic semiconductor is doped with phosphorus, it becomes n-type as Phosphorus is pentavalent.
The 4 electrons in the outer orbit of Phosphorus are shared by the 4 Germanium atoms and the
fifth electron of Phosphorus in the outer orbit is a free electron. But in order that this electron is
completely detached from the parent Phosphorus atom, some energy is to be supplied. This
energy required to separate the fifth electron is called Ionization Energy. The value of ionization
energy for Germanium is 0.012 eV, and in Silicon, it is 0.044 eY. For different impurity materials,
these values will be different, in Silicon and Gennanium. As this energy is small, at room temperature,
we assume that all the impurity atoms are ionized.
2.8 HOLES AND ELECTRONS
In intrinsic semiconductors, n = p = ni. Or the product n x p = ny. In extrinsic semiconductor
say n-type semiconductor practically the electron concentration, 0» OJ. As a result holes, minority
carrier in n type, encounter with free is much larger since n» p. So
when a hole encounter a freeelectrc)\:;, both electrons and holes recombine and the place of hole
is occupied by the free electrons and this probability is much lager since 0 »p. The result is that
both free electron and hole are lost. So the hole density 'p' decreases and also that of electron
density' n' but still 0 » OJ. This is also true in the case of p-type semiconductor p » OJ, n
decreases in acceptor type semiconductor, as a result of recombination, 'p' also decreases but
p >>OJ. It has been observed practically that the net concentration of the electrons and holes
follows the realtion n x p = n7. This is an approximate formula but still valid. Though 'p' decreases
in n-type semiconductor with recombination, '0' also decreases, but 0 » OJ and because of
breaking of covalent bonds, more free electrons may be cr.eated :and '0' increases.
In the case of p-type semiconductor the concentration of acceptor atoms Na» n
l
. Assuming
that all the acceptor atoms are ionized, each acceptor atom contribute at least one hole. So
p » OJ. Holes are the majority carriers and the electrons minority carriers. As p » OJ, the
current is contributed almost all, by holes only and the current due to electrons is negligibly small.
If impurities of both donor type and acceptor type are simultaneously doped in intrinsic
semiconductor, the net result will be, it can be either,p-rype or n-rype depending upon their individual
concentration. To give a specific example, suppose donor atoms concentration is 100 n
l
, and acceptor
atoms concentration in 10 nl" Then No = 0.1 N
I
. The number of electrons combine contributed by
Junction Diode Characteristics 69
ND combine with number of holes contributed by Na. So the net free electrons will be equal to 0.9
No = 90 Ni. Such a semiconductor, can be regarded as n type semiconductor. If NA = ND the
semiconductor remains intrinsic.
2.S.1 INTERSTITIAL ATOMS
Intrinsic or pure semiconductor is practically not available. While doping a semiconductor with
impurities, pure Phosphorous, Arsenic, Boron or Aluminium may not be available. These impurities
themselves will contain some impurities. Commonly found such undesirable impurities are Lithium,
Zinc, Copper, Nickel etc. Sometimes they also act as donor or ac:::eptor atoms. Such atoms are
called as interstitial atoms, except Copper and Nickle other impurities do not affect much.
2.S.2 EFFECTIVE MASS
When quantum mechanism is used to specify the motion of electrons or holes within a crystal,
holes and electrons are treated as imaginary particles with effective masses mp and mn respectively.
This is valid when the external applied field is smaller than the internal periodic fields produced by
the lattice structure.
Most metals and semiconductors are crystalline in structure, i.e., they consist of space
array of atoms in a regular tetrahedral or any other fashion. rile regular pattern of atom
arrangement is called lattice. In the case of metals, in each crystal, the atoms are very close to
each other. So the valency electron of one atom are as much associated with the other atoms as
with the parent atom. In other words, the valance electrons are loosely bound to the parent atom
and the electrons of one atom are shared by another atom. So every such valence
electron has almost zero affinity with any individual atom. Such electrons are free to move within
the body of the metal under the influence of applied electric field. So conductivity of metals is
large.
On the other hand, for a semiconductor also, the valence electrons
of one atom are shared by the other atoms. But these binding forces
are very strong. So the valence electrons are very much less mobile.
Hence conductivity is less; As the temperature is increased, the covalent
bonds binding the valance electrons are broken and electrons made
free to move, resulting in electrical conduction.
Valence Electrons are the outer most electrons orbiting around
the nucleus.
.
•
.
\J
f\
.
.+4
.
.
Free electrons are those valence electrons which are separated
from the parent atom. Since the covalent bonds are broken. Germanium
has 4 valence electrons. Number of electrons is equal to number of Fig 2.6 Covalent bonds
protons. The atom is neutral when no electric field is applied. In the
adjacent figure, the ion is having a charge (circles) with 4 electrons around it. The covalent
bonds are shown by I ines I inking one electron of one atom to the nucleus of other atom (Fig. 2.6).
2.S.3 HOLES AND EXCESS ELECTRONS
When a covalent bond is broken due to thermal agitation, an electron is released and a hole is
created in the structure of that particular atom. The electron so released is calledfree electron or
excess electron since it is not required to complete any covalent bond in its immediate
neighborhood. Now the ion which has lost electron will seek another new electron to fill the
70 Electronic Devices and Circuits
vacancy. So because of the thermal agitation of the crystal lattice, an electron of another ion may
come very close to the ion which has lost the electron. The ion which has lost the electron will
immediately steal an electron from the closest ion, to fill its vacancy. The holes move from the
first ion to the second ion. When no electric field is applied, the motion of free electron is
random in nature. But when electric field is applied, all the free electrons are lined up and they
move towards the positive electrode. The life period of a free electron may be lp-sec to 1
millisecond after which it is absorbed by another ion.
2.9 MASS ACTION LAW
In an intrinsic Semiconductor number of free electrons n = n
l
= No. of holes p = PI
Since the crystal is electrically neutral, nlPI = n ~ .
Regardless of individual magnitudes ofn and p, the product is always constant,
np= n ~
3 -EGO
nj = AT2 e 2KT
.......... ( 2.13 )
This is called Mass Action Law.
2.10 LAW OF ELECTRICAL NEUTRALITY
Let No is equal to the concentration of donor atoms in a doped semiconductor. So when these
donor atoms donate an electron, it becomes positively charged ion, since it has lost an electron. So
positive charge density contributed by them is No. If 'p' is the hole density then total positive
charge density is No + p. Similarly if NA is the concentration of acceptor ions, (say Boron which
is trivalent, ion, accepts an electron, so that 4 electrons in the outer shell are shared by the Ge
atoms ), it becomes negatively charged. So the acceptor ions contribute charge = ( NA + n ).
Since the Semiconductor is electrically neutral, when no voltage is applied, the magnitude of positive
charge density must equal that of negative charge density.
Total positive charge, No + P = Total negative charge (NA + n )
N
o
+ p=NA + n
This is known as Law of Electrical Neutrality.
Consider n-type material with acceptor ion density NA = O. Since it is n-type, number of
electrons is» number ofholes.
So 'p' can be neglected in comparison with n.
nn ::: No. ( Since every Donor Atom contributes one free electron. )
In n-type material, the free electron concentration is approximately equal to the density of
donor atoms.
In n-type semiconductor the electron density nil = No. Subscript n indicates that it is n-type
semiconductor.
But n x p = n
2
n n I
n
2
Pn = The hole density in n-type semiconductor = N ~
n
2
I
PI1 = No
.......... ( 2.14 )
Junction Diode Characteristics
Similarly Hole concentration in p-type semiconductor,
Pp::::: NA·
_ 2
np x Pp - n
l
n
2
71
I
n =-
p NA
.......... ( 2.15 )
Problem 2.14
A specimen of intrinsic Germanium at 300 oK having a concentratiun of carriers of2.5 x 1013/cm3
is doped with impurity atoms of one for every million germanium atoms. Assuming that all the
impurity atoms are ionised and that the concentration of Ge atoms is 4.4 x 10
22
I cm
3
, find the
resistivity of doped material. ( Iln for Ge = 3600 cm
2
/volt-sec).
Solution
The effect of minority carriers (impurity atoms) is negligible. So the conductivity will not appreciably
change.
Conductivity, an = nelln ( neglecting Hole Concentration)
4.4 x 10
22
n = 4.4 x 1022/cm
3
, No = Icm
3
Resistivity,
Problem 2.15
10
6
Iln = 3600 cm
2
I Volt - sec
an = 4.4 x 10
16
x 3600 x 1.6 x 10-
19
= 25.37 mhos I cm
I
Pn = 25.37 = 0.039 O-cm
Determine the conductivity ( a ) and resistivitd: ( p. ) of pure silicon, at 300
0
K assuming that the
concentration of carriers at 300 oK is 1.6 x 10
1
Icm
3
for Si and mobilities as Iln = 1500 cm
2
N-sec;
/lp = 500 cm
2
N-sec.
Solution
Problem 2.16
0'1 = (Iln + /lp) e n
l
= 1.6 x 10-
19
x 1.6 x 10
10
x (1500 + 500)
= 5.12 x 10-
6
u/cm
Pi= 5.12xlO-6 =195,3000-cm
Determine the concentration of free electrons and holes in a sample ofGe at 300 oK which has a
concentration of donor atoms equal to 2 x 10
14
atoms Icm
3
and a concentration of acceptor atoms
= 3 x 10
14
atom lem
3
Is this p-type or n-type Germanium?
Solution
n x p = nr
-Ego
n = AT3/2 e 2kT
I
72
A = 9.64 x 10
14
EG = 0.25 ev
n ~ = 6.25 x 10
26
/cm
3
NA + n = No + P
Total negative charge = Total positive charge
Electronic Devices and Circuits
or p - n = NA - No = (3 - 2) x 10
14
= 10
14
or p = n + 10
14
Then n( n + 10
14
) = 6.25 x 10
26
or n = 5.8 x 10
12
electrons. /cm
3
and p = n + 10
14
= 1.06 x 10
14
holes/cm
3
As p> n, this is p-type semiconductor.
Problem 2.17
Find the concentration of holes and electrons in a p type germanium at 300
o
K, if the conductivity
is 100 n -cm. IIp Mobility of holes in Germanium = 1800 cm
2
I V - sec.
Solution
Problem 2.18
It is p-type p » n.
O"p = P e IIp
0" 100
Pp = ell
p
= 1.6 x 10-
19
x 1800 = 3.47 x 10
17
holes/cm
3
n x p = n7
ni = 2.5 x 10 13 /cm
3
(2.Sx 10
13
f
n = ...l.. ___ ---,-:_=_ = 1.8 x 10
9
electrons/cm
3
3.47x 10
17
( a) Find the concentration of holes and electrons in p-type Germanium at 300
o
K, if
0" = 100 u/cm.
( b) Repeat part ( a ) for n-type Si, if 0" = 0.1 u/cm.
Solution
As it is p-type semiconductor, p» n.
0" = peu
p
cr 100
p= ell
p
= 1.6xlO-
19
x1800
= 3.47 x 10
17
holes/cm3
nxp=n;
3 EGO
ni = AT2 e--
2kT
= 2.5 x 1013/
cm
3
Junction Diode Chara'cteristics 73
p = 3.47 x 1017/
cm
3
n = n ~ = (2.5 x 1013 L = 1.8 x 10
9
electron/m3
p 3.47xl0
17
(b) 0" = ne f..I11 ;
n - 0.1 = 4.81 x 1014/
cm
3
- 1300x 1.6 x 10-
19
= 4.81 x 1014/
cm
3
n, = 1.5 x 10
10
;
P = nf = (1.5x10
lO
L =4.68 x 1011 hole/m3
n 4.81 x 10
14
Pwblem 2.19
A sample of Ge is doped to the extent of 10
14
donor atoms/cm
3
and 7 x 10
13
acceptor atoms/
cm
3
. At room temperature, the resistivity of pure Ge is 60 O-cm. If the applied electric field is
2 V /cm, find total conduction current density.
Solution
For intrinsic Semiconductor,
n = p = nj
O"j = n, e ( J..lp + J..ln )
1
= 60 u/cm.
J..lp for Ge is 1800 cm
2
1V-sec ;
J..ln = 3800 cm
2
1V-sec.
cr I I
nj= e(llp+lln) 60{1.6xIO-19)(3800+1800)
= 1.86 x 1013 electron/cm
3
p x n = nr = ( 1.86 x 10
13
)2
NA + n =N
o
+ P
NA = 7 x 1013/
cm
3
No = 10
14
/cm3
( p - n ) = NA - No = -3 x 10
13
Solving (I) and (2) simultaneously to get p and n,
p = 0.88 x 1013
n = 3.88 x 10
13
J = ( nJ..ln + PJ..lp ). eE
= {( 3.88)( 3800) + (0.88)( 1800)} x 1013. eE
= 52.3 mA/
cm
3
........ ( 1)
........ (2;
74 Electronic Devices and Circuits
Problem 2.20
Determine the concentration of free electrons and holes in a sample of Germanium at 300
0
K
which has a concentration of donor atoms = 2 x 10
14
atoms /cm
3
and a concentration of acceptor
atoms = 3 x 10
14
atoms/cm
3
. Is this p-type or n-type Ge? In other words, is the conductivity due
primarily to holes or electrons?
Solution
n x p =
NA + n = P + No
Solving ( a ) and ( b ) to get nand p,
p - n = NA - No
n
2
p=_.
n
n
2
_n
2
• = (NA - No)
or
n
n.
2
- n
2
= n x ( NA - No )
or n
2
+ n ( NA - No) - n? = 0
This is in the form ax
2
+ bx + c = 0
x -4ac
2 2a
n =_ (NA - No) + i(NA - ND? = 0
2 V 2
Negative sign is not taken into consideration since electron or hole concentration cannot be
negative.
n > 0 and p> O.
NA = 3 x 1014/
cm
30
No = 2 x 1014/
cm
3
nj at 300
0
K = 2.5 x 10
13
/ cm
3
l -EG
n. = AT
2
e
2KT
E=O.72eV
n? = 6.25 x 1026/
cm
3
n can be calculated. Similarly p is also calculated.
n = 5.8 x IOIS/cm
3
p = 10.58 x 1019/
m
3
as p > n, it is p-type semiconductor.
Junction Diode Characteristics 75
Problem 2.21
Calculate the intrinic concentration of Germanium in carries/m
3
at a temperature of320oK given
that ionization energy is 0.75 eV and Boltzman's Constant K = 1.374 x I 0-23J 1
0
K. Also calculate
the intrinsic conductivity given that the mobilities of electrons and holes in pure germanium are
0.36 and O. I 7 m
2
1 volt-sec respectively.
Solution
~ -e EGO
nl=AT2e 2KT
K = Boltzman's Constant in J I oK
K = Boltzman's Constant in eV I oK
_16)(10-
19
xO 75
-21
= 9.64 x 10
21
x (320)3/2. e
2xl
37xl0 . x320
n
l
= 6.85 x 10
19
electrons (or holes)/m3.
In intrinsic semiconductor, n = p = n
l
.
0"1 = enj (f.ln + f.lp)
= 1.6 x 10-
19
x 6.85 x 10
19
(0.36 + O. 17) = 5.797 u/m
Problem 2.22
Determine the resistivity of intrinsic Germanium at room temperature
Solution
T = 300 oK
A=9.64xI0
21
E = 0.75 eY.
3 -EGO
n
l
= AT 2e ~ = 2.5 x 10
19
electrons (or holes) 1m
3
.
f.ln = 0.36m2 I V - sec
f.lp = 0.17 m
2
I V - sec
O"j = enj( f.ln + f.lp )
= 1.6 x 10-
19
x 2.5 x 10
19
(0.35 + 0.17) = 2.13 u/m
1 1
p = - = -- = 0.47 O-m
(J 2.13
2.11 THE FERMI DIRAC FUNCTION
N( E ) = Density of states.
i.e., The number of states per ev per cubic meter ( number of statesle V 1m
3
)
The expression for
N(E) = Y E
Y
' where y is a constant.
76 Electronic Devices and Circuits
47t
Y = h3 (2m)3/2 (1.6 x 10-
19
)3/2 = 6.82 x 10
27
m = Mass of Electron in Kgs
h = Planck's Constant is Joule-sees.
The equation for./{ E ) is called the Fermi Dirac Probability Function. It specifies the
fraction of all states at energy E (eV) occupied under conditions of thermal equilibrium. From
Quantum Statistics, it is found that,
where
./(E) = (E-E
F
)
l+e kT
k = Boltzmann Constant, eVfK
T= Temp oK
EF = Fermi Level or Characteristic Energy
........ ( 2.16 )
The momentum of the electron can be uncertain. Heisenberg postulated that there is always
uncertainty in the position and momentum of a particle, and the product of these two uncertainities
is of the order of magnitude of Planck's constant 'h'.
If ~ p is the Uncertainty in the Momentum of a particle, ~ n is the uncertainty in the position
of a particle
~ p x ~ x ~ h.
2.11.1 EFFECTIVE MASS
When an external field is applied to a crystal, the free electron or hole in the crystal responds, as
if its mass is different from the true mass. This mass is called the Effective Mass of the electron
or the hole.
By considering this effective mass, it will be possible to remove the quantum features of the
problem. This allows us to use Newton's law of motion to determine the effect of external forces
on the electrons and holes within the crystal.
2.11.2 FERMI LEVEL
Named after Fermi, it is the Energy State, with 50% probability of being filled if no forbidden
band exists. In other words, it is the mass energy level of the electrons, at OaK.
IfE = E6
./(E) = ~ From Eq.( 2.17 ) .
If a graph is plotted between ( E - EF ) and./{ E), it is shown in Fig. 2.7
At T = OaK, if E > EF then,./{E) = o.
That is, there is no probability of finding an electron having energy> EF at T = 0° K. Since
fermi level is the max. energy possessed by the electrons at Oak . ./(E) varies with temperature as
shown in Fig. 2.7.
Junction Diode Characteristics 77
f( E)
+ _____ --1._....;;:=_-...; __ eV
-1.0
( E- E,)
+1.0
Fig. 2.7 Fermi level variation with temperature.
2.11.3 UNCERTAINTY PRINCIPLE
This was proposed by Heisenberg. The measurement of a physical quantity is characterized in an
essential way by lack of precision.
2.11.4 THE INTRINSIC CONCENTRATION
I
j( E ) = 1+ e(E-E
F
)/KT
e(E-EF)/KT
I-j(E)= l+e(e-EF)/KT :::e-(Ef-E)/KT
e(E-E
F
)/KT[l +e(E-Er)/KT l' ::: e-(E,-E)/KT
Fermi function for a hole = I - j( E ).
( EF - E ) » KT for E .:::: Ev
the number of holes per m
3
in the Valence Band is,
Ev
p = f Y{Ev - xdE
where
Similarly
-00
= N
v
x e-(EF-Ey)/KT
3

.......... ( 2.22 )
So Fermi Level EF is close to Conduction Band Ec in n-type semiconductor.
Similarly for p-type material,
P=NA
Junction Diode Characteristics
But
N A = e-(EF-Ev)/KT
.. N
v
Taking Logarithms,
or
In NA = _(EF-Ev )
N
v
KT
N
KTx In---A- = Ev - EF
Ny
EF = EY+KTXlnR I
NA=N
y
Fermi Level is close to Valance Band E v in p-type semiconductor.
Problem 2.25
83
.......... ( 2.23 )
In n type silicon, the donor concentration is I atom per 2 x 10
8
silicon atoms. Assuming that the
effective mass of the electron equals true mass, find the value of temperature at which, the fermi
level will coincide with the edge of the conduction band. Concentration of Silicon = 5 x 10
22
atom!
cm3.
Solution
Donor atom concentration I atom per 2 x 10
8
Si atom.
5 x 10
22
atoms/cm
3
Silicon atom concentration
5 x 10
22
.. No = 2xl0
8
= 2.5 x 10
14
/
cm
3.
For n-type, Semiconductor,
EF = Ee - KT In ( Ne )
No
If EF were to coincide with E
e
, then
Ne = No'
No = 2.5 x 10
14
/cm
3
.
3
Nc ~ 2{ 2X:;KT}'
h = Plank's Constant; K = Boltzman Constant
mn the effective mass of electrons to be taken as = mE
3
N = 2{2
X
3.14
X
9.IX.IO-
31
xKXT}2
e h 2
3
= 4.28 x 10
15
T2
84 Electronic Devices and Circuits
3
4.28 x 10
15
T2 = 2.5 x 10
14
T=O.14°K
2.12 TOTAL CURRENT IN A SEMICONDUCTOR
2.12.1 DRIFT CURRENT IN AN N-TvPE SEMICONDllCTOR
Within a semiconductor, intrinsic or impure, because of the thermal energy, covalent bonds are
broken and electrons ar,d holes move in random directions. These collide with lattices, get deflected
and move in a different direction, till they collide with another carrier. Such a random motion is
defined as mean free pat" length 'I', the distance a carrier travels between collisions and the
I
average time between collisions 't'. The average velocity of motion v = t. Over a period oftime
which is » '1', average movement is zero or net current is zero.
But when electric field is applied all the electrons are aligned in a particular direction and
move towards the positive electrode and holes in the opposite direction. The resulting current is
called Drift Current.
Let the semiconductors be 'n' type. Now using a battery, electric field is applied, under the
influence of the electric field, all the free electrons move towards the positive electrode and enter
the metal of the positive electrode Fig. 2.10. The donor atoms have thus lost their free electrons.
So the donor atoms !lear the positive electrode pull electrons from the electrode, exactly equal in
number to the free electrons which have entered the electrode. So the semiconductors remains
electrically neutral. The voltage applied results in voltage drop across semiconductor. If the
battery is removed, the number of free electrons and holes is same as before the application of
field, since semiconductor has taken equal number of electrons from the positive electrode.
,...., ,-.
+ -
0) 0)
0
n - type semi conductor
-
"-.I
.
~
+ : r-1 -
Fig. 2.10 Drift current in n-type semiconductor.
When the free electron contributed by phosphorus donor atom goes into the positive electrode,
the donor atom loses one electron. Therefore it will pull one more electron from the positive
electrode and hence the number of free electrons in the semiconductors remains the same.
Junction Diode Characteristics 85
Problem 2.26
In p-type Silicon, the acceptor concentration corresponds to 1 atom per 10
8
Silicon atoms. Assume
that m = 0.6m. At room temperature, how far from the edge of the valence band is the Fermi
level?" Is EF above or below Ev? The concentration of Silicon atoms is 5 x 1022 atoms/cm
3
.
Solution
Concentration of Silicon atoms = 5 x 10
22
/cm
3c
.
Because doping is done at 1 atom per 10
8
Silicon Atoms.
where
or
5 x 1 022
N = = 5 x 10
14
/cm
3
A 10
8
- ((27tm
p
KT)% 1
N
v
- 2 h2
K = Boltzman's Constant in eVfK; K = Boltzman constant in J// oK.
h = Planck's Constant = 6.62 x 10-
34
J-sec
K = 1.38 x 10-
23
J/oK
3
= 4.82 x 10
15
x ( m p ~ x T3/2
m
m = 0.6 m (given)
p
T=300oK
N
v
= 4.82 x 10
15
(0.6 x 300)3/2 = 1.17 x 10
19
/cm
3
EF - Ev = K T In ( ~ : )
(
1.17 X 10
19
)
= 0.026 In . 5 x 10
14
=0.026 x 10
= 0.26 ev
EF - Ev = 0.26ev
EF is above Ev'
86 Electronic Devices and Circuits
Problem 2.27
In p-type Ge at room temperature of 300 oK, for what doping concentration will the fermi level
coincide with the edge of the valence bond? Assume ~ = 0.4 m.
p
Solution
when
3
N
v
=4.82 x 1015(:YXT
312
=4.82 x 10
15
(0.4)3/2(300)312
= 6.33 x 10
18
•
Doping concentration N A = 6.33 x 10
18
atoms/cm
3
•
Problem 2.28
If the effective mass of an electron is equal to thrice the effective mass of a hole, find the distance
in·electron volts (ev) offermi level in as intrunsic semiconductor from the centre of the forbidden
bond at room temperature.
Solution
For Intrunsic Semiconductor,
If
then
Hence EF will be at the centre of the f o ~ bidden band. But if m "t: m . EF will be away from
the centre of the forbidden band by P n
KT Nc kT mn
2 In. N
v
= ~ 2.1n mp
(
-]3/2
21tmnKT
N = 2 2
C n
(
- J3/2
21tmpkT
N =2 2
V n
3
= "4 x 0.026 In (3) = 21.4 meV
Junction Diode Characteristics
2.12.2 DRIFT CURRENT IN P-TVPE SEMICONDliCTOR
The mechanism is the same to as
explained above. The holes in the
acceptor type semiconductor moves
towards the negative electrode and enter -
into it, pulling out one electron from the
p-Type
'--' '-"
!--
negative electrode from the acceptor
atoms ( Fig. 2.11 ), the hole has moved
away, i.e. it has acquired an electron.
So electrical neutrality or of its original
~ - - - - - - - - - - ~ : I I ~ - - - - - - - - - - - - - - - - ~
condition is disturbed. This results in a Fig. 2.11 Drift current in p-type semiconductor
electrons from the acceptor atom being
87
pulled away. These free electrons enter the positive electrode. The acceptor atoms having lost
one electron steal another electron from the adjoining atom resulting in a new hole. The new holes
created thus drift towards negative electrode.
2.12.3 DIFFlJSION CllRRENT
This current results due to difference in the concentration gradients of charge carriers. That is,
free electrons and holes are not uniformly distributed all over the semiconductor. In one particular
area, the number of free electrons may be more, and in some other adjoining region, their number
may be less. So the electrons where the concentration gradient is more move from that region to
the place where the electrons are lesser in number. This is true with holes also.
Let the concentration of some carriers be as shown in the Fig 2.12. The concentration of
carriers is not uniform and varies as shown along the semiconductor length. Area A I is a measure
of the number of carriers between XI and X2. Area A2 is a measure of the number of carriers
between X2 and X3. Area Al is greater than Area A
2
. Therefore number of carriers in area Al is
greater than the number of carriers in A
2
. Therefore they will move from Al to A
2
. If these
x, x, x,
Area A,
Distance of ( along the Semiconductor)
Fig. 2.12 Diffusion current.
88 Electronic Devices and Circuits
charge carriers are free electrons, then electron current directions is from C to A (.: Electrons
move from A to C). If these are holes then current direction is from A to C. This current is called
diffusion current. It is independent of any applied field. So in semiconductors, at room temperature
itself, though no electric field is applied, if the device is connected in a circuit, there will be very
small current flowing through, which is cal1ed Diffusion Current.
Let Ix be the distance travel1ed by each electron between two successive col1isions., i.e. I x
is the mean free path. Let tx secs be the corresponding time taken.
Because of the thermal agitation, the carriers in areas AI and A2 have equal probability to
move from left to right or from right to left. But because the number of carrier in AI is greater
than in A
2
, it results in net diffusion current. During time t
x
, half the electrons from AI move to A2
and half the electrons from A2 move to AI' The net current is due to the difference of number of
electrons from AI to A2
=0.5 X(lx ::)XdX
Hence Diffusion Current = ((lx)2 X :: )( t: )
1/ I dn
/2 e x x X V n X dx
Ix
where V x is the Average Velocity = t and e = charge of an electron.
x
Hence Diffusion Current per Sq. meter of cross section is
dn
in = e x On X dx
On = Diffusion current constant in m
2
/sec
= Y:! Ix Vx
:: is concentration gradient of electrons in Number of carriers / m.
If diffusion current is caused by holes, the equation of diffusion current is
. dp
Ip = Q x Dp x dx
For Ge, at room temperature
On = 93 x 10-
4
m
2
/sec
Dp = 44 x 10-
4
m
2
/sec.
Diffusion electrons density I
n
is given by the expression
dn
J = + e On x -
n dx
positive sign is lIsed since :: is negative and direction of current is opposite to the movement
of electrons.
Junction Diode Characteristics 89
where Dn is called Diffusion Constant for electrons. It is in m
2
/sec. :: is concentration gradient.
D and 11 are interrelated. It is given as,
On Op
- = - =V
T
~ n ~ p
where
KT T
V
T
= -;- = \\,600
.......... ( 2.24 )
where V
T
is volt equivalent. of temperature. At room temperature, 11 = 39D.
The thermal energy due to temperature T is expressed as electrical energy in the form
ofVoIts.
2.12.4 TOTAL CURRENT
Both Potential Gradient and Concentration Gradient can exist simultaneously within a
Semiconductor. Since in such a case the total current.is the sum of Drift Current and
Diffusion Current.
dp
Jp = e IIp p. E -e Dp . dx
Drift Current ellppE
dp
Diffusion Current eDp dx
Similarly the net electron current is
dn
In = e Iln nE + e Dn dx
S i n ~ e Diffusion hole current is J
p
dp
Jp = --eDp dx
p decreases with increase in x. So : ~ is negative. Negative sign is used for J
p
, so that, J
p
will
be positive in the positive x direction. For electrons
dn
I n = + e . Dn . dx
since the electron current is opposite to the directions of conventional current.
There exists a concentration gradient in a semiconductor. On account of this, it results in
Diffusion Current. If you consider p-type semiconductor holes are the majority carriers. So the
resulting Diffusion Current Density J
p
is written as
dp
Jp = - e x Dp x dx
where Dp is called diffusion constant for holes. Since p the hole concentration is decreasing
. I dp . S .. d· h . fi J
WIt 1 x, dx IS -ve. 0 -ve sIgn IS use In t e expressIon or p.
90 Electronic Devices and Circuits
dn
Similarly for electrons also the expression is similar and the slope is - dx . But since the
electron current is opposite to the conventional current,
(
dn) dn
J = e.D
n
x- +exD
n
-
n dx dx
2.13 EINSTEIN RELATIONSHIP
D, the Diffusion Coefficient and ~ are inter related as
I ~ ~ ~ ~ V T I
where Vr is volt equivalent of temperature.
KT T
V
r
= ~ = 11,600
K = 1.6 x 10-
19
JfK
K = 8.62 x 10-
5
evfK
K = 1.381 x 10-
23
JfK
D D
~ p = -p = -p- =39D
Vr 0.026 p
~ p = 39D
p
~ = 39D
n
or ~ = 39D
Therefore values ofDp and Dn for Si and Ge can be determined.
Problem 2.29
.......... ( 2.25 )
.......... ( 2.26 )
.......... ( 2.27 )
Determine the values ofDp and Dn for Silicon and Germanium at room temperature.
Solution
For Germanium at room temperature,
Dn = ~ x V
r
= 3,800 x 0.026 = 99 cm
2
/sec
pp = ~ p x V
r
= 1800 x 0.026 = 47 cm
2
/sec
For Silicon, Dn = 1300 x 0.026 = 34 cm
2
/sec
Dp = 500 x 0.026 = 13 cm
2
/sec
2.14 CONTINUITY EQUATION
Thus Continuity Equation describes how the carrier density in a given elemental volume of
crystal varies with time.
If an intrinsic semiconductor is doped with n-type material, electrons are the majority carriers.
Electron - hole recombination will be taking place continuously due to thermal agitation. So the
concentration of holes and electrons will be changing continuously and this varies with time as well
as distance along the semiconductor. We now derive the differential. equation which is based on
the fact that charge is neither created nor destroyed. This is called Continuity Equation.
Junction Diode Characteristics
Consider a semiconductor of area A, length
dx (x +. dx - x = dx) as shown in Fig. 2.l3. Let
the average hole concentration be 'p'. Let Ep is a
factor ofx. that is hole current due to concentration
is varying with distance along the semiconductor.
Let Ip is the current entering the volume at x at
time t, and (Ip + dIp) is the current leaving the
volume at (x + dx ) at the same instant of time '1'.
So when only Ip colombs is entering, ( Ip + dIp)
colombs are leaving. Therefore effectively there
is a decrease of ( Ip + dIp - Ip ) = dIp colombs per
second within the volume. Or in other words, since
more hole current is leaving than what is entering,
we can say that more holes are leaving than the
no. of holes entering the semiconductor at 'x'.
If dIp is rate of change of total charge that is
91
x'--_-;
dIp = d( n:q)
Fig 2.13 Charge flow in semiconductor
dI
_P gives the decrease in the number of holes per second with in the volume A x dx. Decrease
q
in holes per unit volume ( hole concentration) per second due to Ip is
But
dIp 1
---x
Axdx q
dI
:. = Current Density
1
q
dJ
p
x-
dx .
But because ofthermal agitation, more number of holes will be created. If Po is the thermal
equilibrium concentration ofholes,( the steady state value reached after recombination ), then, the
increase per second, per unit volume due to thermal generation is,
p
g=-
Lp
Therefore, increase per second per unit volume due to thermal generation,
.&
g=
'p
But because of recombination of holes and electrons there will be decrease in hole concentration.
The decrease
92 Electronic Devices and Circuits
Charge can be neither created nor destroyed. Because of thermal generation, there is
increase in the number of holes. Because of recombination, there is decrease in the number of
holes. Because of concentration gradient there is decrease in the number of holes.
So the net increase in hole concentration is the algebraic sum of all the above.
a p P" - P 1. aJ p
-=----x-
at 't
p
q ax
dp
Partial derivatives are used since p and Jp are functions of both time t and distance x. dt
gives the variation of concentration of carriers with respect to time 't'.
If we consider unit volume of a semiconductor ( n-type ) having a hole density Pn, some
holes are lost due to recombination. Ifpno is equilibrium density, (i.e., density in the equilibrium
condition when number of electron = holes ).
The recombination rate is given as Pn - Pno . The expression/or the time rate 0/ change
't
in carriers density is called the Continuity Equation.
R b
" R dp
ecom .natIOn rate = dt
Life time of holes in n-type semiconductors
AP Pn-Pno
't =- =
P R dp/dt
or
dp = [pn -PnoJ
dt 't
p
where Pn is the original concentration of holes in n-type semiconductors and Pno is the concentration
after holes and electron recombination takes place at the given temperature. In other word Pno is
the thermal equilibrium minority density. Similarly for a p-type semiconductors, the life time
of electrons
_ np -n
po
't
n
- dxl dt
2.15 THE HALL EFFECT
If a metal or semiconductor carrying a current I is placed in a perpendicular magnetic field
B, an electric field E is induced in the direction perpendicular to both I and B. This
phenomenon is known as the Hall Effect. It is used to determine whether a semiconductor
is p-type or n-type. By measuring conductivity 0; the mobility !1 can be calculated using
Hall Effect.
In the Fig. 2.14 current I is in the positive X-direction and B is in the positive Z-direction. So
a force will be exerted in the negative V-direction. If the semiconductor is n-type, so that current
is carried by electrons, these electrons will be forced downward toward side 1. So side 1 becomes
negatively charged with respect to side 2. Hence a potential V H called the Hall Voltage appears
between the surface 1 and 2.
Junction Diode Characteristics 93
2
..............
. . .
z
Fig 2.14 Hall effect.
In the equilibrium condition, the force due to electric field intensity 'E', because of Hall
effect should be just balanced by the magnetic force or
or
But
where
or
or
eE = B ev
v Drift Velocity of carriers in m / sec
B = Magnetic Field Intensity in Tesla ( wb/m
2
)
E = Bv
E = VH/d
V H = Hall Voltage
d = Thickness of semiconductors.
J = nev or J = pv
P = charge density.
J Current Density ( Amp / m
2
)
I
J=-
ro.d
ro = width of the semiconductor; rod = cross sectional area
I = current
I
J = Current Density = rod
E = VH/d
V
H
= Ed
.......... ( a )
94
But
Electronic Devices and Circuits
E = Bv .......... From Equation { a )
V H = B x v x d But v = J/p
BJ.d I
= -- But J=-
p rod
B.I.d B.I
V
H
=--=-
p.ro.d pro
~
. I
V
H
=-
pro
.......... ( 2.28 )
If the semiconductor is n-type, electrons the majority carriers under the influence of electric
field will move towards side 1, side 2 becomes positive and side 1 negative. If on the other hand
terminal 1 becomes charged positive then the semiconductor is p-type.
or
and
p = n x e ( For n - type semiconductor)
p = p x e ( for p-type semiconductor)
p = Charge density.
B.I
V --
H - pro
B.I
P = VH.ro
.......... ( 2.29 )
I
The Hall Coefficient, RH is.defined as RH = p. Units of RH are m
3
I coulombs
If the conductivity is due primarily to the majority carriers conductivity, cr = ne/J. in n-type
semiconductors.
But
or
n.e = p = charge density.
cr=pX/J.
1
- =R
H
P
1
cr=--X/J.
RH
VH·ro
/J. = RH x cr = -- x cr
B.I
Junction Diode Characteristics 95
We have assumed that the drift velocity 'v' of all the carriers is same. But actually it will not
be so. Due to the thermal agitation they gain energy, their velocity increases and also collision with
other atoms increases. So for all particles v will not be the same. Hence a correction has to be
1 31t
made and it has been found that satisfactory results will be obtained if R H is taken as 8p .
.......... ( 2.30 )
8
Multiply RH by 31t. Then it becomes Modified Hall Coefficient. Thus mobility of
carriers ( electrons or holes) can be determined experimentally using Hall Effect.
The product Bev is the Lorentz Force. because of the applied magnetic field B and the drift
velocity v. So the majority carriers in the semiconductors, will tend to move in a direction
perpendicular to B. But since there is no electric field applied in that particulars direction, there
will develop a Hall voltage or field which just opposes the Lorentz field.
So with the help of Hall Effect, we can experimentally detemline
1. The mobility of Electrons or Holes.
2. Whether a given semiconductor is p-type or n-type (from the polarity
of Hall voltage V H)
Problem 2.30
The Hall Effect is used to determine the mobility of holes in a f-type Silicon bar. Assume the bar
resistivity is 200,000 O-an, the magnetic field B
z
= 0.1 Wblm and d = w = 3mm. The measured
values of the current and Hall voltage are 1 OmA and 50 mv respectively. Find IIp mobility of holes.
Solution
B = 0.1 Wb I m
2
(or Tesla)
V
H
= 50 mv.
1= 10 rnA;
p = 2 x 10
5
0 - cm ;
d = w = 3mm = 3 x 10-
3
meters
B.I 0.1 x lOx 10-
3
1
RH VH·w 50xlO-
2
x3xl0-
3
= 150 =0.667.
1
Conductivity = p = 2 x 105 x I 0-2 - 2000 mhos I meter.
J-l = cr x RH
I 1
J-lp = 0.667 x 2000 = 750 cm
2
1 V - sec
96 Electronic Devices and Circuits
2.16 SEMICONDUCTOR DIODE CHARACTERISTICS
If ajunction is formed usingp-lype and n-type semiconductors, a-diode is realised and it has the
properties of a rectifier. In this chapter, the volt ampere characteristics of the diodes, electron-
hole currents as a function of distance from the junction and junction capacitances will be studied.
2.16.1 THEORY OF p-n JUNCTION
Take an intrinsic Silicon or Germanium crystal. If donor ( n-type ) impurIties are diffused from
one point and acceptors impurities from the other, a p-njunction is formed. The donor atoms will
donate electrons. So they loose electrons and become positively charged. Similarly, acceptor
atoms accept an electron, and become negatively charged. Therefore in the p-njunction on the p-
side, holes and negative ions are shown and on the n-side free electrons and positive ions are
shown. To start with, there are only p-type carriers to the left of the junction and only n-type
carriers to the right of the junction. But because ofthe concentration gradient across the junction,
holes are in large number on the left side and they diffuse from left side to right side. Similarly
electrons will diffuse to the right side because of concentration gradient.
(a)
(b)
(c)
(d)
(e)
eee e®

,
, Potential Barrier for:
Holes !
x
-l\j Potential Barrier Cor Electron,
iii -+x
Fig 2.15. Potential distribution in p-n junctioh diode.
Junction Diode Characteristics 97
Because ofthe displacement of these charges, electric field will appear across the junction.
Since p-side looses holes, negative field exists near the junction towards left. Since n-side looses
electrons, positive electric field exists on the n side. But at a particular stage the negative field on
p-side becomes large enough to prevent the flow of electrons from n-side. Positive charge on
n-side becomes large enough to prevent the movement of holes from the p-side. The charge
distribution is as shown in Fig. 2.15 ( b). The charge density far away from the junction is zero,
since before all the holes from p-side move to n-side, the barrier potential is developed. Acceptor
atoms near the junction have lost the holes. But for this they would have been electrically neutral.
Now these holes have combined with free electrons and disappeared leaving the acceptor atom
negative. Donor atoms on n-side have lost free electrons. These free electrons have combined
with holes and disappeared. So the region near the junction is depleted of mobile charges.
This is called depletion region, space charge region or transition region. The thickness of
this region will be of the order offew microns
1 micron = 10-
6
m = 10-4cm.
The electric field intensity near the junction is shown in Fig. 2.15 ( c). This curve is the
integral of the density function p. The electro static potential variation in the depletion region is
f
dV
shown in Fig. 2.15 ( d ). E = - dx' This variation constitutes a potential energy barrier against
further diffusion of holes across the barrier.
When the diode is open circuited, that is not connected in any circuit, the hole current must be
zero. Because of the concentration gradient, holes from the p-side move towards n-side. So, all
the holes from p-side should move towards n-side. This should result in large hole current flowing
even when diode is not connected in the circuit. But this will not happen. So to counteract the
diffusion current, concentration gradient should be nullified by drift current due to potential barrier.
Because of the movement of holes from p-side to n-side, that region (p-region ) becomes negative.
A potential gradient is set up across the junction such that drift current flows in opposite
direction to the diffusion current. So the net hole current is zero when the diode is open circuited.
The potential which exists to cause drift is called contact potential or diffusion potential. Its
magnitude is a few tenths of a volt ( 0.01 V ).
2.16.2 p-n JlINCTION AS A DIODE
The p-n Junction shown here forms a semiconductor device called DIODE. Its symbol is
A ~ I K. A is anode. K is the cathode. It has two leads or electrodes and hence the name
Diode. If the anode is connected to positive voltage terminal of a battery with respect to cathode,
it is called Forward Bias, ( Fig. 2.16 (a». 'fthe anode is connected to negative voltage terminal
ofa battery with respect to cathode, it is called Reverse Bias, (Fig. 2.16 (b».
A
K A
K
+ - - +
( a ) Forward bias ( b ) Reverse bias ( c ) p-n junction forward bias
Fig 2.16
98 Electronic Devices and Circuits
2.16.3 OHMIC CONTACT
In tpe above circuits, external battery is connected to the diode. But directly external supply
cannot be given to a semiconductor. So metal contacts are to be provided for p-region and n-region.
A Metal-Semiconductor Junction is introduced on both sides of p-n junction. So these must be
contact potentials across the metal-semiconductor junctions. But this is minimized by fabrication
techniques and the contact resistance is almost zero. Such a contact is called ohmic contact. So
the entire voltage appears across the junction of the diode.
A K
+ -
Fig 2.17 Ohmic contacts.
2.17 THE P-N JUNCTION DIODE IN REVERSE BIAS
Because of the battery connected as shown, holes in ( Fig. 2.16 (b)) p-rype and electrons in n-
rype will move away from the junction. As the holes near the junction in p-region they will move
away from the junction and negative charge spreads towards the left of the junction. Positive
charge density spreads towards right. But this process cannot continue indefinitely, because to
have continuous flow of holes from right to left, the holes must come from the n-side. But n-side
has few holes. So very less current results. But some electron hole pairs are generated because
of thermal agitation. The newly generated holes on the n-side will move towards junction. Electrons
created on the p-side will move towards the junction. So there results some small current called
Reverse Saturation Current. It is denoted by 1
0
, 10 will increase with the temperature. So the
reverse resistance or back resistance decreases with temperature. 10 is of the order of a few
~ A . The reverse resistance of a diode will be of the order of Mil For ideal diode, reverse
resistance is 00.
I he same thing can be explained in a different way. When the diode is open circuited, there
exists a barrier potential. Jfthe diode is reverse biased, the barrier potential height increases by
a magnitude depending upon the reverse bias voltage. So the flow of holes from p-side to n-side
and electrons from n-side to p-side is restricted. But this barrier doesn't apply to the minority
carriers on the p-sides and n-sides. The flow ofthe minority carriers across the junction results
in some current.
2.18 THE P-N JUNCTION DIODE IN FORWARD BIAS
When a diode is forward biased, the potential barrier that exists when the diode is open circuited,
is reduced. Majority carriers from p-side and n-side flow across the junction. So a large current
results. For ideal diode, "the forward resistance R
f
= O. The forward current If' will be of the
order ofmA (milli-amperes).
Junction Diode Characteristics 99
A K
+ -
Fig 2.18 Diode in forward bias.
2.1S.1 FORWARD ClJRRENT
Ifa large forward voltage is applied (Fig. 2.19), the current must increase. Ifthe barrier potential
across the junction is made zero, infinite amount of current should flow. But this is not practically
possible since the bulk resistance of the crystal and the contact resistance together will limit the
current. We may see in the other sections that when the diode is conducting, the voltage across it
remains constant at V y cut in voltage. If the applied voltage is too large junction breakdown
will occur.
'--__ A-7"i+:r-
I
_
K
_---._'--J
V
Fig 2.19 Forward biasing.
2.19 BAND STRUCTURE OF AN OPEN CIRCUIT p-D JUNCTION
Whenp-rype and n-rype semiconductors are brought into intimate contactp-njunction is formed.
Then the fermi level must be constant throughout the specimen. If it is not so, electrons on one side
will have higher energy than on the other side. So the transfer of energy from higher energy
electrons to lower energy electrons will take place till fermi level on both sides comes to the same
level. But we have already seen that in n-type semiconductors, EF is close to conduction band Een
and it is close to valence band edge Ey on p-side. So the conduction band edge of n-type
semiconductor cannot be at the same level as that of p-rype semiconductor Hence, as shown, the
energy band diagram for a p-n junction is where a shift in energy levels Eo is indicated.
EG Energy gap in eV
EF Fermi energy level
Eo Contact difference of potential
Een Conduction Band energy level on the n-side.
Ecp Conduction Band energy level on the p-side.
Eyn Valence Band energy level on the n-side.
Eyp Valence Band energy level on the p-side.
100
I I
I Space charge I
Electronic Devices and Circuits
P region region n region
I
;EG
, .2 1
--
1
"2 EG
Fig 2.20 Band structure of open circuited diode.
E
If a central line 2
G
is taken, the shift in energy levels is the difference between the two
E
central lines 2
G
of the two semiconductors.
But
But
Eo = Eep - Een = Evp - Evn·
EG
EI = 2- (EF - Evp)
EG
E2 = 2 - (Een - E
F
)
EI + E2 = EG - EF + Evp - Een + EF
EG = Eep - Evp
EI + E2 = Eep - Evp - Een + Evp·
Eep - Een = Eo
E} + E2 = Eo·
This energy Eo represents the potential energy barrier for electrons. The contact difference
of potential
.......... (1 )
Junction Diode Characteristics
and Een - EF = 1;2 ( EG ) - E2 = ) - E2
Adding (1) and (2),
or
( EF - Evp) + ( Een - EF ) = EG - E) - E2
( E) + E2 ) = EG - ( Een - EF ) - ( EF - Evp )
(E) + E2 ) = Eo
(E - E ) = KT In[Nc)
en F No
No = Donor Atom Concentration NJm
3
.
N A = Acceptor Atom Concentration NJm
3
•
( EF - Evp) = KT
Eo = KT x x KT
n, c v n,
The energy is expressed in electron volts eV.
K is Botlzman's Constant in eV 1
0
K = 8.62 x 10-
5
eV I oK
101
.......... (2)
.......... (2.32)
Therefore, Eo is in eV and Vo is the contact difference potential in volts Vo is numerically
equal to Eo. In the case of n-type semiconductors, nn = No. (Subscript 'n' indicates electron
concentration in n-type semiconductor)
n
2
=n xp =N xp
1 n nOn
nn=N
o
n = Intrinsic Concentration
1
np = Electron Concentration in p-type semiconductor
nn = Electron Concentration in n-type semiconductor
Pp = Hole Concentration in p-type semiconductor
Pn = Hole Concentration in n-type semiconductor
n
2
p =_1 and
n No
n
2
n = _I and
p NA
n
2
= n x p
1 n n
2

Current
Total Current I
....................... ..
x = 0 Distance
Fig 2.21 Current components in a p-n junction.
Junction Diode Characteristics 103
When a forward bias is applied to the diode, holes are injected into the n-side and electrons to the
p-side. The number of this injected carriers decreases exponentially with distance from the
junction. Since the diffusion current of minority carriers is proportional to the number of carriers,
the minority carriers current decreases exponentially, with distance. There are two minority currents,
one due to electrons in the p-region Jnp, and due to holes in the n-region Jpn. As these currents
vary with distance, they are represented as Jpn(.y;).
Electrons crossing from n to p will constitute current in the same direction as holes crossing
from p to n. Therefore, the total current at the junction where x = 0 is
1 = Ipn(O) + Inp(O)
The total current remains the same. The decrease in Jpn is compensated by increase in Jnp
on the p-side.
Now deep into the p-region ( where x is large) the current is because of the electric field
( since bias is applied) and it is drift current Jpp of holes. As the holes approach the junction, some
of them recombine with electrons crossing the junction from n to p. So Jpp decreases near the
junction and is just equal in magnitude to the diffusion current Jnp. What remains of Jpp at the
junction enters the n-side and becomes hole diffusion current Jpn in the n-region. Since holes are
minority carriers in the n-side, Jpn is small and as hole concentration decreases in the n-region,
Jpn also exponentially decreases with distance.
In a forward biased p-n junction diode, at the edge of the diode on p-side, the current is
hole current (majority carriers are holes). This current decreases at the junction as the junction
approaches and at a point away from the junction, on the n-side, hole current is practically zero.
But at the other edge of the diode, on the n-side, the current is electron current since electrons are
the majority carriers. Thus in a p-n junction diode, the current enters as hole current and leaves
as electron current.
2.21 LAW OF THE JUNCTION
Ppo Thermal Hole Concentration onp-side
Pno Thermal equilibrium hole concentration on n side
Ppo PnoeVO/VT .......... (1)
where V 0 is the Electrostatics Barrier Potential that exists on both sides of the Junction. But the
thermal equilibrium hole concentration on the p-side
where
Pn (0) e(Vo -V)/VT
Hole concentration on n-side near the junction
Applied forward bias voltage.
This relationship is called Boltzman's Relationship.
or
Equating (1) and (2).
Pn (0) e(VO-V)/VT = Pno X e VO/VT
Vo _ Vo +
Pn(O) = P
no
X e V, V, V,
Pn(o)=Pnox e
V/VT
.......... (2)
104 Electronic Devices and Circuits
Therefore, the total hole concentration in 'n' region at the junction varies with applied forward
bias voltage V as given by the above expression.
This is called the Law of the Junction.
Pn(O) Pn(O) - Pno
P
V/VT _ P
no e no
2.22 DIODE CURRENT EQUATION
The hole current in the n-side Ipn(x) is given as
Aex Op -xjL
Jpn(x) = L Pn (0) e P
p
But Pn(O) Pno (e v /v
r
-1)
Ipn(O)
AexOp (v/v)
L x Pno\e 1-1
p
Diffusion coefficient of holes
Diffusion coefficient of electrons.
. . e -xjLp at x = 0 is 1.
.......... ( 2.32 )
Similarly the electron current due to the diffusion of electrons from n-side to p-side is
obtained from the above equation itself, by interchanging nand p.
AexOn
.. Inp(O) = x n (e v/v
1
-1)
Ln po ~
The total diode current is the sum of Ipn (0) and Inp(O)
or I I = 10 (e
v
/
v1
-1) I
.......... ( 2.33 )
AeDp AeOn
where 10 = -- x Pno+ L x npo
Lp n
In this analysis we have neglected charge generation and recombination. Only the current
that results as a result of the diffusion of the carriers owing to the applied voltage is considered.
Reverse Saturation Current
1= 10 x (e
V
/
VI
-1)
This is the expression for current I when the diode is forward biased. If the diode is
reverse biased, V is replaced by -Y. V
T
value at room temperature is - 26 mY. If the reverse
-v
bias voltage is very large, VT is very small. So it can be neglected.
e
1 = -10
Junction Diode Characteristics
where
10 will have a small value and 10 is called the Reverse Saturation Current.
AeDpPno AeDnnpo
10 = L +
p
In n-type semiconductor,
°n=N
D
But n x p = n.
2
n n 1
In p-type semiconductor,
n
2
Pp=NA :. np= N
1
A
Substituting these values in the expression for 1
0
,
10 =Ae [ ~ + Dn Jx n
7
LpN
o
Ln.N
A
1
n 2 = A T3 -EG/KT
I 0 e
EG is in electron volts = e.V G' where V G is in Volts.
But
EGO
n? =Ao T3 e-
KT
EGo = V
G
· e
KT
- = Volt equivalent of Temperature V T'
e
105
For Germanium, Dp and Dn decrease with temperature and o? increases with T. Therefore,
temperature dependance of 10 can be written as,
10 = K
J
T2 e-VG/VT
For Germanium, the current due to thermal generation of carriers and recombination can be
neglected. But for Silicon it cannot be neglected. So the expression for current is modified as
(l \
I = 10 l e 11 V
T
- 1 j
where n = 2 for small currents and n = 1 forM large currents.
2.23 VOLT-AMPERE CHARACTERISTICS OF A P-N JUNCTION DIODE
The general expression for current in the p-njunction diode is given by
(l \
1=10 le
11VT
-lj.
11 = 1 for Germanium and 1 or2 for Silicon. For Silicon, I will be less than thatforGermanium.
V
T
= 26 mY.
106 Electronic Devices and Circuits
If V is much larger than V T' 1 can be neglected. So I increases exponentially with forward
bias voltage Y. In the case of reverse bias, if the reverse voltage -V» V T' then e -v /VT can be
neglected and so reverse current is -10 and remains constant independent ofY. So the characteristics
are as shown in Fig. 2.22 and not like theoretical characteristics. The difference is that the
practical characteristics are plotted at different scales. If plotted to the same scale, (reverse and
forward) they may be similar to the theoretical curves. Another point is, in deriving the equations
the breakdown mechanism is not considered. As V increases Avalanche multiplication sets in.
So the actual current is more than the theoretical current.
1,(mA)
---:-4 - - - ~ t - - + ' ' ' ' - - - - : - - V" (Volts)
lillA)
Fig 2.22 V-I Characteristics of p-n junction diode.
CUT IN VOLTAGE Vy
In the case of Silicon and Germanium, diodes there is a Cut In or Threshold or OffSet or Break
Point Voltage, below which the current is negligible. It's magnitude is O.2V for Germanium and
O.6V for Silicon ( Fig. 2.23 ).
i
Silicon
)
O.2V O.6V V F (Volts)
Fig 2.23 Forward characteristics of a diode.
Junction Diode Characteristics 107
2.23.1 DIODE RESISTANCE
V
The static resistance (R) of a diode is defined as the ratio of I of the diode. Static resistance
varies widely with V and I. The dynamic resistance or incremental resistance is defined as the
dV
reciprocal of the slope of the Volt-Ampere Characteristic ill. This is also not a constant but
depends upon V and I.
2.24 TEMPERATURE DEPENDANCE OF P-N JUNCTION DIODE CHARACTERISTICS
The expression for reverse saturation current 10
= __ p_+ n X 2
[
D D J
10 Ae L N L N nj
n 2 a T3
I
p 0 n· A
n 2 = A T3 e-EG0/KT
I 0
Dp decreases with temperature.
10 a T2
or 10 = KTm e-VGO/llVT
where V G is the energy gap in volts. (EG in eV)
For Germanium, 11 = 1, m = 2
For Silicon, 11 = 2, m = 1.5
10 = KTm e-VGO/llVT
Taking In, (Natural Logarithms) on both sides,
-VGO
In 10 = In( K ) + m x In (T) --
llV
T
Differentiating with respect to Temperature,
m
~ x dl
o
0 + m _ (_ Voo x J..)
10 dT T 11Vr T
dl
o
x-
10 dT
- value is negligible 7
T
~ dl
o
x-
10 dT
Experimentally it is found that reverse saturation current increases::: 7% /oC for both
Silicon and Germanium or for every lOoC rise in temperature, 10 gets doubled. The reverse
saturation current increases if expanded during the increasing portion.
108 ElectrQnic Devices and Circuits
25V+4---
·0
IJ1A
.I'-I1---IOflA
Fig 2.24 Reverse characteristics of a p-n junction diode.
Reverse Saturation Current increases 7% JOC rise in temperature for both Silicon and
Germanium. For a rise of IOC in temperature, the new value of 10 is,
10 = (1+ I ~ O ) l o
= 1.071
0
,
For another degree rise in temperature, the increase is 7% of(1.07 1
0
),
Therefore for 10 °C rise in temperature, the increase is (1.07)10 = 2.
Thus for every 10°C rise in temp Iofor Silicon and Germanium gets doubled.
2.25 SPACE CHARGE OR TRANSITION CAPACITANCE C
T
When a reverse bias is applied to a p-njunction diode, electrons from the p-side will move to the
n-side and vice versa. When electrons cross the junction into the n-region, and hole away from
the junction, negative charge is developed on the p-side and similarly positive charge on the n-
side. Before reverse bias is applied, because of concentration gradient, there is some space
charge region. Its thickness increases with reverse bias. So space charge Q increases as
reverse bias voltage increases.
Q
r"' __
'-' - V
But
Therefore, Incremental Capacitance,
C
T
= I : ~ I
where IdQI is the magnitude of charge increase due to voltage dV. It is to be noted that there is
negative charge on the p-side and positive charge on n-side. But we must consider only its magnitude.
dQ
Current I = dt
Junction Diode Characteristics
Therefore, ifthe voltage dV is changing in time dt, then a current will result, given by
dV
1= C
T
x dt
109
This current exists for A.C. only. For D.C. Voltage is not changing with time. For D.C.
capacitance is open circuit.
The knowledge of C
T
is important in considering diode as a circuit element. C T is called
Transition region capacitance or space charge capacitance or barrier capacitance, or
depletion region capacitance. This capacitance is not constant but depends upon the reverse
bias voltage V. If the diode is forward biased, since space charge:.:::: O,(this doesn't exist). It will
be negligible. C
T
is of the order of 50 pf etc.
ALLOY JVNCTJON
Indium is trivalent. If this is placed against n-type Germanium, and heated to a high temperature,
indium diffuses into the Germanium crystal, a pnjunction will be formed and for such ajunction
there will be abrupt change from acceptor ions on one side to donor ions on the other side. Such
a junction is called Alloy Junction or Diffusion Junction. In the figure, the acceptor ion
concentration NA and donor atom concentration, No is shown in Fig. 2.25. There is sudden
change in concentration levels. To satisfy the condition of charge neutrality,
ex NA x Wp = e x No x W
n
.
p-type
Charge Density
r
NA
I
-ve I •
•• -1-.
IE
I W
I
P
I
I
• • •
r-- +ve
•
•
•
•
D
•
•
•
I
I_
I
I
I
I
I
I
I
I
I
!(
W
W
n
n-type
----+
x
Fig 2.25 Abrupt p-n junction.
IfNA «No, then W
p
» W
n
. In practice the width of the region, W n will be very small. So
it can be neglected. So we can assume that the entire barrier potential appears across the
p-region just near the junction.
110 Electronic Devices and Circuits
Poisson's Equation gives the relation between the charge density and potential.
d
2
V eNA
dx2 = -E-
It is
where E is the permitivity of the semiconductor
dV eNA
-=--x
dx E
e.N
A
x
2
V= -- X-,
E 2
At x = w P' V = VB the barrier potential. W p = W.
IV B = ~ x W2
1
.......... ( 2.34 )
The value of W depends upon the applied reverse bias V. If V 0 is the contact potential,
VB = V 0 - V where V is the reverse bias voltage with negative sign.
where
So as V increases, W also increases and VB increases
Wa JVB
If A is the area of the junction, the charge in the width W is
Q=exNAxWxA
But
or
W xA=Volume
e x N A = Total charge density
Cr = I ~ I =exNAxAx I ~ ~ I
Q JexN
A
J2E VB xA
e.N
A
2
V=--xW
B 2E
dW
dV
Rxf
E Y:
= y: v- 2
2 ex NA B
~ V B = W ~ e ~ ~ A
Junction Diode Characteristics
dW
dV
~ &
~ ~ ~ x ~ e x N A x W
E I
---x-
exNA W
E
111
.......... ( 2.35 )
This expression is similar to that of the Parallel Plate Capacitor.
2.26 DIFFUSION CAPACITANCE, CD
When a p-n junction diode is forward biased, the junction capacitance will be much larger than
the transition capacitance C
r
When the diode is forward biased, the barrier potential is reduced.
Holes from p-side are injected into the n-side and electrons into the p-side. Holes which are the
minority carriers in the n-side are injected into the n-side from the p-region. The concentration
of holes in the n-side decrease exponentially from the junction. So we can say that a positive
charge is injected into the n-side from p-side. This injected charge is proportional to the applied
forward bias voltage 'V'. So the rate of change of injected charge 'Q' with voltage 'V' is called
the Diffusion Capacitance CD' Because of CD total capacitance will be much larger than C
T
in
the case of forward bias, (CD is few mF (2mF.)) C
T
value will be a few pico farads.
DERIVATION FOR CD
Assume that, the p-side is heavily doped compared to n-side. When the diode is forward biased,
the holes that are injected into the n-side are much larger than the electrons injected into the
p-side. So we can say that the total diode current is mainly due to holes only. So the excess
charge due to minority carriers will exist only on n-side. The total charge Q is equal to the area
under the curve multiplied by the charge of electrons and the cross sectional area A of the diode.
P n(O) is the Concentration ofholes/cm3. Area is in cm
2
, x in cm,
1
Carrier
Concentration
p(O)
n
p(O}
n
= Hole concetration in n-region at x = 0
= Thermal Equilibrium Concentration of Holes.
x=O
---+.x
Fig 2.26 Carrier concentration variation.
112
..
We know that
or
Electronic Devices and Circuits
00
Q =
fAeP
n
(O)e-
x/Lp
dx
0
00
=
J -x/Lp d
Ae Pn(O) e . x
0
Q AePn(O) [ - Lp [0-1] ]
AeLpPn(O)
C =
dQ dPn(O)
- =AeL x
D
dv p
dv
Ipn( x )
+ exAxDpxPn(O)xe-
X/LP
Lp
AeDpPn(O)
Ipn(O)= 1=--'---
Lp
Lp x I / AeD
p
Lp dl
--x-
AeDp dv
Lp
AeDp x g
.......... (1)
.......... (2)
where g is the conductance of the diode.
where
Substitute equation (2) in (1).
CD ~ Ae x Lp x [ A ~ ~ , ) x g
L2
= i xg
Dp
The lifetime for holes 't
p
= 't is given by the eq.
L2
't= -p
Dp
Dp = Diffusion coefficient for holes.
Lp = Diffusion length for holes.
D = cm
2
/sec
p
CD = 't x g.
Junction Diode Characteristics
where
But diode resistance
11,
V
T
r---
- I
11 = 1 for Germanium
11 = 2 for Silicon
I
g=-
11,
V
T
1..
113
C = - (3)
D 11,
V
T .......•..
CD is proportional to 1. In the above analysis we have assumed that the current is due to
holes only. So it can be represented as COP' If the current due to electron is also to be considered
then we get corresponding value of COn' The total diffusion capacitance = COP + C
Dn
. Its value
will be around 20pF'.
Co =. x g
or r x Co =.
r x CD is called the time constant of the given diode. It is of importance in circuit applications.
Its value ranges from nano-secs to hundreds of micro-seconds.
Charge control description of a diode:
But
Q = A x e x Lp x Pn(O)
AeDpPn(O)
1= ---'---
Lp
Q
L = A x e x Pn(O)
p
I=QxD/L2
p p
L 2/Dp = •
p
Q
1= -
•
2.27 DIODE SWITCHING TIMES
When the bias of a diode is changed from forward to reverse or viceversa. the current takes
definite time to reach a steady state value.·
2.27.1 FORWARD RECOVERY TIME (T )
FR
Suppose a voltage of 5V is being applied to the diode. Time taken by the diode to reach from 10%
to the 90% of the applied voltage is called as the forward recovery time t
fT
. But usually this is very
small and so is not of much importance. This is shown in Fig. 2.26.
2.27.2 DIODE REVERSE RECOVERY TIME (t
rr
)
When a diode is forward biased, holes are injected into the 'n' side. The variation of concentration
of holes and electrons on n-side and p-side is as shown in Fig. 2.27. P no is the thermal equilibrium
concentration of holes on n-side. P
n
is the total concentration of holes on 'n '-side.
114 Electronic Devices and Circuits
i
v
)
i x=o
Time
Fig 2.27 Rise time. Fig 2.28 Carrier concentration in reverse bias.
2.27.3 STORAGE AND TRANSITION TIMES
Suppose the input given to the circuit is as shown in Fig. 2.29 (i). During 0 - t\, the diode is
forward biased. Forward resistance of the diode is small compared to R
L
. Since all the voltage
drop is across RL itself, the voltage drop across the diode is small.
V
F
IF = RL
This is shown in Fig 2.29 (ii) and voltage across the diode in Fig 2.29
(iii) up to t\. Now when the forward voltage is suddenly reversed, at t = t\, because of the
Reverse Recovery Time, the diode current will not fall suddenly. Instead, it reverses its direction
V
and ~ R: (RL is small compared to reverse resistance of the diode). At t = ~ , the equilibrium
level of the carrier density at the junction takes place. So the voltage across the diode falls slightly
but not reverses and increases to V R after time t3' The current also decreases and reaches a
value = reverse saturation current 10 .
The interval time (t\ - ~ ) for the stored minority charge to become zero is called Storage
Time t
s
. The time ( ~ - t
3
) when the diode has normally recovered and the reverse current reaches
10 value is called Transition Time tt. These values range from few mill i-seconds to a few
micro-seconds.
Junction Diode Characteristics
( i )
t.
P.- Pno
junction
Ir=S
R,

( ii )

v

I t. Minority
Forward I Carrier
B" I I
las I Storage I
( iii )
liS
-V R ------..L---.!!!I!.'!..---..L.....;:o ........ --
I+--- t t
I S I
! ! Transition
intorval
.1
1<
t
rr
Fig 2.29 Storage and Transition times.
Problem 2.31
(a) For what voltage will the reverse current in p-njunction Germanium diode reach 90%
of its saturation value at room temperature?
(b) If the reverse saturation current is 10 JlA, calculate the forward currents for a voltage
of 0.2, and O.3V respectively"
116 Electronic Devices and Circuits
Solution
T
(a) V
T
= 11,600 = 0.026V at room temperature. Or it is 26mV
V
T
is volt equivalent of temperature. T is in
KT
V
T
= -;-
It is the Thermal Energy expressed in equivalent electrical units of Volts.
T) = I, for Germanium Diode.
or
(b)
Problem 2.32
0.9 10 - 10 I J
V =-O.017V
For V = 0.2, 1 = 10 (e200/26 -1) = 21.85 rnA
0.2V =200mV
For V=0.3,I=IO(e
I
1.
52
_1)=1.01A
Find the value of (i) D.C resistance and a.c resistance of a Germanium junction diode, if the
temperature is 25°C and 10= 20J.IA with an applied voltage of 0.1 V
Solution
1 = 10 (e
XYr
-1)
T = 273 + 25 = 298°K
For Germanium, T) = 1
T (273+25)
V T = 11 600 = 11 600 = 0.026V
, ,
1=20x e
VT
-1 =0.916mA
(
0.1 J
V 0.1
r
DC
= I = 0.196 x 10-
3
= 109.20
dV I1V
r AC = dI = 111 =?
Junction Diode Characteristics
Problem 2.33
1 dI
r =-=-
AC g dv
ac
dI
dv Vr
rAC = 26.30
117
= 38.1 x 10-
3
0
Find the width of the depletion layer in a germanium junction diode which has the following
specifications Area A = 0.001 cm
2
, <I
n
= Imhos I cm, <J
p
= 100 mhos I cm, fln = 3800 cm
2
/sec
flp = 1800 cm
2
/sec.
Solution
Permitivity of Germanium,
E = 16 x 8.85 x 10-
14
F/cm
n? = 6.25 x 10
26
T = 300
0
K.
Applied reverse bias voltage = 1 V.
J2E.VB
W=
e.N
A
In this formula, in the denominators, N A is used since in the expression we have assumed
thatp-side of the p-njunction is heavily doped. lfn-side is heavily doped, it would be N
D
•
V
B
= Total barrier potential Applied reverse bias voltage
+ the contact difference of potential
(Vo ).
V
B
= V
R
+V
So first V 0 should be calculated.
nn·pp
Vo=KT In. -2-
nj
KT = 0.026 eV
W=
2xI6x8.85 x lO-
14
X(0.35+I) -4
- - - - ~ - - ' - : - = - - ~ = 0.083 x 10 cm
1.6x 10-
19
x 3.5 x 10
17
118 Electronic Devices and Circuits
Problem 2.34
Calculate the dynamic forward and reverse resistance of a p - n junction diode, when the applied
voltage is 0.25V for Germanium Diode. 10 = lilA and T = 300
0
K.
Solution
10 = lilA
T = 300
0
K
V
f
=0.25V
V
r
= 0.25V
1 = 10 (e
YVr
- t)
For Germanium, II = I
Dynamic Forward Resistance :
V is positive
v
I dI 10 v
-=-=-eT_O
r
f
dV V
T
I x 10--6 0.25
= . e0
026
0.026
1
rf = 0.578 mhos
r
f
= 1.734 n
Dynamic Reverse Resistance :
dI 10 _I<
=-=
V
T
. e Vr
rr dV
I x 10-
6
-0.25
--
0.026
eO.0
26
1
- = 2.57 x 10-
9
mhos
rr
rr = 2.57xlO-9 =389.7 Mn
In practice rr is much smaller due to surfac.e leakage current.
2.28 BREAK DOWN MECHANISM
There are three types of breakdown mechanisms in semiconductor devices.
1. Avalanche Breakdown 2. Zener Breakdown 3. Thermal Breakdown
Junction Diode Characteristics 119
2.28.1 AVALANCHE BREAKDOWN
When there is no bias applied to the diode, there are certain number of thermally generated carriers.
When bias is applied, electrons and holes acquire sufficient energy from the applied potential to
produce new carriers by removing valence electrons from their bonds. These thermally generated
carriers acquire additional energy from the applied bias. They strike the lattice and impart some
energy to the valence electrons. So the valence electrons will break away from their parent atom
and become free carriers. These newly generated additional carriers acquire more energy from
the potential (since bias is applied). So they again strike the lattice and create more number of
free electrons and holes. This process goes on as long as bias is increased and the number offree
carriers gets multiplied. This is known as avalanche multiplication, Since the number of carriers
is large, the current flowing through the diode which is proportional to free carriers also increases
and when this current is large, avalanche breakdown will occur.
2.28.2 ZENER BREAKDOWN
Now if the electric field is very strong to disrupt or break the covalent bonds, there will be sudden
increase in the number of free carriers and hence large current and consequent breakdown. Even
if thermally generated carriers do not have sufficient energy to break the covalent bonds, the
electric field is very high, then covalent bonds are directly broken. This is Zener Breakdown. A
junction having narrow depletion layer and hence high field intensity will have zener breakdown
effect. (== 10
6
V 1m). If the doping concentration is high, the depletion region is narrow and will
have high field intensity, to cause Zener breakdown.
2.28.3 THERMAL BREAKDOWN
If a diode is biased and the bias voltage is well within the breakdown voltage at room temperature,
there will be certain amount of current which is less than the breakdown current. Now keeping the
bias voltage as it is, if the temperature is increased, due to the thermal energy, more number of
carriers will be produced and finally breakdown will occur. This is Thermal Breakdown.
In zener breakdown, the covalent bonds are ruptured. But the covalent bonds of all the
atoms will not be ruptured. Only those atoms, which have weak covalent bonds such as an atom
at the surface which is not surrounded on all sides by atoms will be broken. But if the field
strength is not greater than the critical field, when the applied voltage is removed, normal covalent
bond structure will be more or less restored. This is Avalanche Breakdown. But if the field
strength is very high, so that the covalent bonds of all the atoms are broken, then normal structure
will not be achieved, and there will be large number offree electrons. This is Zener Breakdown.
In Avalanche Breakdown, only the excess electron, loosely bound to the parent atom will become
free electron because of the transfer of energy from the electrons possessing higher energy.
2.29 ZENER DIODE
This is a p-n junction device, in which zener breakdown mechanism dominates. Zener diode is
always used in Reverse Bias.
Its constructional features are:
1. Doping concetration is heavy on p and n regions of the diode, compared
to normal p-n junction diode.
2. Due to heavy doping, depletion region width is narrow.
120
3.
Electronic Devices and Circuits
V
Due to narrow depletion region width, electric field intensity E = d
V
= ;; will be high, near the junction, of the order of J06V/m. So Zener
Breakdown mechanism occurs.
In normal p-n junction diode, avalanche breakdown occurs if the applied voltage is very
high. The reverse characteristic ofa p-njunction diode is shown in Fig. 2.29.
When the Zener diode is reverse biased, the current flowing is only the reverse saturation
current 10 which is constant like in a reverse biased diode. At V = V z' due to high electric
field ), Zener breakdown occurs. Covalent bonds are broken and suddenly the number of free
electrons increases. So I
z
increases sharply and V z remains constant, since, I
z
increases through
Zener resistance R
z
decreases. So the product V z = R
z
. I
z
almost remains constant. If the input
voltage is decreasea, the Zener diode regains its original structure. (But if Vi. is increased much
beyond V z' electrical breakdown ofthe device will occur. The device looses Its semiconducting
properties and may become a short circuit or open circuit. This is what is meant by device
breakdown. )
Applications
1. In Voltage Regulator Circuits
2. In Clipping and Clamping Circuits
3. In Wave Shaping Circuits.

In
Fig 2.29 Reverse characteristic of a Zener diode.
2.30 THE TUNNEL DIODE
In an ordinary p-n junction diode the doping concentration of impurity atoms is 1 in 10
8
. With this
doping, the depletion layer width, which constitutes barrier potential is 5 J.l V. If the concentration of
the impurity atoms is increased to say 1 in 1 03( This corresponds to impurity density of 10
19
1m
3
),
the characteristics of the diode will completely change. Such a diode is called Tunnel Diode.
This was found by Esaki iri 1958.
2.30.1 TUNNELING PHENOMENON
Barrier Potential VB :
"
Junction Diode Characteristics
or
J2E,V
B
W=
e.N
A
.
where W = Width of Depletion Region in J.L
N A = Impurity Concentration N
o
/m
3
.
121
So, the width of the. junction barrier varies inversely as the square root of impurity
concentration. Therefore as N A increases W decreases.
Therefore, in tunnel diodes, by increasing NA' W can be reduced from 5J.L to 0.01J.L. According
to classical mechanics, a particle must possess the potential which is at least equal to, or greater
than, the barrier potential, to move from one side to the other. When the barrier width is so thin as
0.0 I J.L, according to Schrodinge equation, there is much probability that an electron will penetrate
when a forward bias is applied to the diode, so that potential barrier decreases below E . The
n-side levels must shift upward with respect to those on the p-side. So there are states
in the conduction band of the n material, which are at the same energy level as allowed empty
states in the valence band of the p-side. Hence electrons will tunnel from the n-side to the p-side
giving rise to forward current. As the forward bias is increased further, the number of electrons
on n-side which occupy the same energy level as that vacant energy state existing on p-side,
also increases. So more number of electrons tunnel through the barrier to empty states on the left
side giving rise to peak current Ip. If still more forward bias is applied, the energy level of the
electrons on the n-side increases, but the empty states existing on the p-side, reduces. So the
tunneling current decreases. In addition to the Quantum Mechanical Tunneling Current, there
is regular p-njunction injection current also. The magnitude ofthis current is considerable only
beyond a certain value of forward bias voltage.
Therefore, the current again starts beyond V v' The graph is shown in Fig. 2.30.
r ....... .
I Ip
F
rnA
VRE-
Volts
Fig 2.30 V - I Characteristics 0/ a Tunnel diode.
Ip = Peak current
Iv = Valley current
V F = Peak forward voltage
50mV
122 Electronic Devices and Circuits
2.30.2 CHARACTERISTICS OF A TUNNEL DIODE
For small forward voltage (ie., Vp ~ 50mV for Ge), forward resistance is small ~ 5n and so
dI
current is large. At the Peak Current I
p
' corresponding to voltage V P' dV is zero. If V is beyond
V p' the current decreases. So the diode exhibits Negative Resistance Characteristics between
Ip and Iv called the Valley Current. The voltage at 'Yhich the forward current again equals Iv is
called as peak forward voltage V v. Beyond this voltage, the current increases rapidly.
The symbol for tunnel diode is shown in Fig. 2.31(a). Typical values ofa tunnel diode are:
Vp = 50 mY, V
v
=350 mY,
Ip
-=8
Iv
Ip = 10 rnA, Negative resistance Rn = -30n.
Series Ohmic resistance Rs = 1 n. The series inductance Ls depends upon t ~ e lead length
and the geometry of the diode package. Ls ~ 5 nH. Junction capacitance C= 20 pF. Its circuit
equivalent is shown in Fig. 2.31(b).
C
(a) (b)
Fig 2.31 (a) Symbol of Tunnel Diode ( b ) Equivalent Circuit
Advantages :
1. Low Cost
2. Low Noise
3. Simplicity
4. High Speed
5. Low Input Power
6. Environmental Immunity.
Disadvantages :
-R
N
1. Low output voltage swings, even for small voltage, while the current goes to large
values. So the swing is limited.
2. Circuit design difficulty, since it is a two terminal device and there is no isolation
between input and output.
Junction Diode Characteristics 123
Applications :
1. As a high frequency oscillator (GHz)
2. As a fast switching device. Switching time is in nano-seconds. Since tunneling is a
quantum mechanical phenomenon, there is no time lag between the application of
voltage and consequent current variation. So it can be used for high frequencies.
If the barrier width is ~ 3A 0, the probability that electrons will tunnel through the barrier is
large. The barrier width will be ~ 3Ao, when impurity doping concentration is::: 1019/cm
3
. If
width ~ 0.01 Jl electrons will tunnel without the application offield. But if width ~ 3A 0, very small
applied voltage is sufficient for the electron to tunnel.
The two conditions to be satisfied for tunneling phenomenon to take place are:
Necessary Condition :
1. The effective depletion region width near the junction must be small, of the order of
3A ° by heavy doping.
Sufficient Condition:
2. There must be equivalent empty energy states on the p-side corresponding to energy
levels of electrons on the n-side, for these electrons to tunnel from n-side to p-side.
When the tunnel diode is reverse biased, there will be some empty states on n-side
corresponding to filled states on the p-side. So electrons will tunnel through the barrier from the
p side to n side [since when the diode is reverse biased the energy levels on n side will decrease].
As bias voltage is increased the number of electrons tunneling will increase, so forward current
increases. When the diode is forward biased, electrons from n-side will increase, and tunnel
through the barrier to p-side. This is called quantum mechanical current. Apart from this, normal
p-n diode current will also be there. Beyond valley voltage V v' it is normal diode forward current.
2.31 VARACTOR DIODE
Barrier of transition capacitance C
r
varies with the value of reverse bias voltage. The larger the
reverse voltage, the larger the W.
ExA
W
So C
r
of a p-n junction diode varies with the applied reverse bias voltage.
Diodes made especially for that particular property of variable capacitance with bias are
called Varactors, Varicaps or Voltacaps. These are used in LC Oscillator Circuits.
The Symbol is shown in Fig. 2.33
Varactor diodes are used in high frequency circuits.
In the case of abrupt junction,
EA
C= W
But W a (vy;,
C a (Vt
Y2
In the case oflinearly graded junction,
W a (V)1I3
C a (Vy1/3
--. - . . . - . J ~ r
•
Fig 2.33 Varactor Diode
124 Electronic Devices and Circuits
The variation of impurity atom concentration with co for different types of junctions is shown
in Fig. 2.34.
Hyper Abrupt Linearly Graded

i = 1m Sin a between 0 5 a 7t
i = 0 between 7t 5 a 5 27t
Rr is the forward resistance of the diode. RL is the load resistance
3.2.2 READING OF DC AMMETER
If a D.C. Ammeter is connected in the rectifier output circuit, what reading will it indicate? Is it
the peak value or will the needle oscillate from 0 to maximum and then to 0 and so on, or will it
indicate average value? The meter is so constructed that it reads the average value.
Area of the curve
By definition, average value = ------
Base
For Half wave rectified output, base value is 21t for one cycle,
138 Electronic Devices and Circuits
n 1m I In
I = - JimSina da = -. -Cos
a
O
DC 21t 0 2n
I = ~ [ I + I] = ~
DC 21t 1t
Upper limit is only 1t, because the signal is zero from 1t to 2n. The complete cycle is from 0 to 21t.
3.2.3 READING OF A.C AMMETER
An A.C. ammeter is constructed such that the needle deflection indicates the effective or RMS
current passing through it.
Effective or RMS value of an A.c. quantity is the equivalent D.C. value which produces the
same heating effect as the alternating component. If some A.C. current is passed through a
resistor, during positive and negative half cycles, also because of the current, the resistor gets
heated, or there is some equivalent power dissipation. What is the value ofD.C. which produces
the same heating effects as the A.c. quantity? The magnitude of this equivalent D.C. is called the
RMS or Effective Value of A.c.
By definition
Inns =
I 2n 2
- J (lmSina) da
2n 0
20r {I -c
2
os2a }da 1m
=&
a 2n Sin2a 2n
+--
2 0 4 0
. 1m ~
RMS value of a slOe wave = ili x ~ 2 + 0
1m J; 1m
= .fiJ; = J2 = 0.707 1m
RMSValue
Form Factor = A V I
verage a ue
1 1t
For a sine wave, 1 = 0.6361 = 2 x - f(lm Sina)da
average m 21t 0
1m
Inns = J2 = O.7071
m
(Ji) O.707I
m
Fonn factor ~ ('; 1 ~ 0.
6361
m
= I.lI
3.2.4 PEAK INVERSE VOLTAGE
For the circuit shown, the input is A.C. signal. Now during the positive half cycle, the diode
conducts. The forward resistance of the diode R
f
wi II be small.
The voltage across the diode v = i x R
f
i = 1m Sin a o :s a:s 1t.
v = 1m RfSin a o :s a :S.1t.
Rectifiers, Filters and Regulators
139
v
AC Input -f==""-" --'\----,---
\J ____ (J. - wt
-Vm " ••• " ••• " ••• ~
Fig 3.5 Fig 3.6 Half Wave Rectifier Circuit
Since R
f
is small, the voltage across the diode V during positive half cycle will be small, and
the waveform is as shown. But during the negative half cycle, the diode will not conduct. Therefore,
the current i through the circuit is zero. So the voltage across the diode is not zero but the voltage
of the secondary of the transformer VI will appear across the diode (.: effectively the diode is
across the secondary of the transformer.)
v = V m Sin a 1t ::: a ::: 21t.
The waveform across the diode is Fig. 3.7.
v
I,nR,sina
o
1t
--0.
V",sino
Fig 3.7 Voltage Waveform across the Diode
So the D.C. Voltage that is read by a D.C. Voltmeter is the average value.
1 [1t 21t 1
V = - PmRf·Sino. do. + JVmSino. do.
DC 21t 0 1t
V m = 1m (R
f
+ Rc)
I V D C ~ -':RLI
.......... ( 3.1 )
If we connect a CRO across the diode, this is the waveform that we see is as shown in
Fig. 3.8. So in the above circuits the diode is being subjected to a maximum voltage of V m.
It occurs when the diode is not conducting. Hence it is called the Peak Inverse Voltage ( PIV )
140 Electronic Devices and Circuits
3.2.5 REGULATION
The variation of D.C output voltage as a function of D.C load current is called 'regulation'.
YNoLoadYoltage - YFullLoad 100
% Regulation = x %
YFullLoad
For an ideal power supply output voltage is independent of the load or the output in voltage
remains constant even if the load current varies, like in Zener diode near breakdown. Therefore,
regulation is zero or it should be low for a given circuit.
EXPRESSION FOR V
DC
THE OUTPUT DC VOLTAGE
For halfwave rectifier circuit (Fig. 3.8), IDe the average value is:
I 21t
= - Ji.da
2n 0
1 1t I
h
-cot
= -flmsina.da =--Dl
2n 1t
Fig 3.8 HWR current output
o
=!nl
1t
.......... ( 3.2
But RL = Load Resistance
R
f
= Forward Resistance of the Diode.
But
Rectifiers, Filters and Regulators 141
Adding and Subtracting R
f
,
Vm·(R
L
+R
f
-R
f
)
V
De
= 1t{R
f
+ R
L
)
Vm
IDe = 1t{R
f
+ R
L
)
IDe is determined by R
L
. Hence V
De
depends upon R
L
.
I V
De
= ~ - IDe x Rfl .......... ( 3.3 )
V
This expression indicates that V De is -...!!!.. at no load or wlten tlte load current is zero, and
1t
it decreases with increase in IDe I inearly since R
f
is more or less constant for a given diode. The
larger the value of R
f
, the greater is the decrease in V DC with IDC' But, the series resistance of
the secondary winding of the transformer Rs should also be considered.
For a given circuit of half wave rectifier, if a graph is plotted between V De and IDe the slope
of the curve gives ( R
f
+ Rs ) where R
f
is the forward resistance of diode and Rs the series
resistance of secondary of transformer.
REGULATION FOR HWR
The regulation indicates how the DC voltage varies as a function of DC load current. In general,
the percentage of regulation for ideal power supply is zero. The percentage of regulation is defined as
V
NL
- V
FL
% Regulation = x 100
V
FL
As we know that,
Vm
V DC = - - I
DC
. R
f
1t
Vm Vm
V
NL
= - and V
FL
= - - IDe' R
f
1t 1t
Vm Vm I R
---+ DC' f
1t 1t
:. % Regulation = -'-'--=-=---'-'---- x 100
Vm I R
-- DC' f
1t
IDc-Rf
VITI I R
-- DC' f
1t
R
f
% Regulation = R x 100
L
x 100
142
Electronic Devices and Circuits
Suppose for a given rectifier circuit, the specifications are 15V and 100 rnA, i.e., the no load
voltage is 15V and max load current that can be drawn is 100 mAo If the value of ( Rr + Rs ) = 25 0
then the percentage regulation of the circuit is:
No Load Voltage = 15V
Drop across Diode = 1m ( R
f
+ Rs )
Rs = Transformer Secondary Resistance
Max. Voltage with load = 15V - ( 1m x ( Rr + Rs ) )
= 15 - 100 rnA x 250
= 15 - 2.5 volts = 12.5V
Percentage Regulation =
3.2.6 RIPPLE FACTOR
15-12.5
12.5
2.5
x 100 = 12.5 x 100::: 20%.
The purpose of a rectifier circuit is to convert A.C. to D.C. But the simple circuit shown before
will not achieve this. Rectifier converts A.C. to unidirectional flow and not D.C. So filters are
used to get pure D.C. Filters convert unidirectional flow into D.C. Ripple factor is a measure of
the fluctuating components present in rectifier circuits.
RMS Value of alternating components of the waveform
Ripple factor, y = ----------=-----'----------
A verage Value of the waveform
I'nns V'rms
y =--=--
I
DC
V
DC
I:ms and V ~ s denote the value of the A.C components of the current and voltage in the
output respectively. While determining the ripple factor of a given rectifier system experimentally,
a voltmeter or ammeter which can respond to high frequencies ( greater than power supply frequency
50 Hz) should be used and a capacitor should be connected in series with the input meter in order
to block the D.C. component. Ripple factor should be small. (Total current i = 1m Sin rot according
to Fourier Series, only A.C. is sum of D.C. and harmonics ).
We shall now derive the expression for the ripple factor. The instantaneous current is given
by i' = i - loc-
i is the total current. In a half wave rectifier, some D.C components are also present.
Hence A.C component is,
i' = ( i-IDe) [ ( Total current - IDe) ]
RMS value is l:ms =
Rectifiers, Filters and Regulators
Now,
I' =
nns
_1_ 2]i
2
dex = Square of the rms value of a sine wave by definition.
21t 0
=(Inni
1 21t
- J i.dex = The average value or D.C value loc.
21t 0
loc is constant. So taking this term outside,
I 2 21t I
DC f dex = de [21t]
21t 0 21t
I ~ s = ~ ( l n n s Y -2Ioc2 + loc
2
The rms ripple current is
I ~ s = ~ ( I n n s Y -J
OC
2
I'rms ~ ( l n n s Y - loc
2
y=-=
Ide IOC
Ripple factor,
143
y=
[
l rms )2 -I
loc
.......... ( 3.4 )
This is independent of the current waveshape and is not restricted to a half wave configuration.
If a capacitor is used to block D.C. and then Inns or V nns is measured,
Then,
I' = ~ I 2 I
nns nns DC
and Ioc = 0 (blocked by Capacitor) output waveform
For Half Wave Rectifier Circuit (HWR),
I = 1m sin ex
1[
I = I = _1_ JIm sinex.dex = ~
av DC 21t 0 1t
144
Ripple Factor
Total current
Electronic Devices and Circuits
1 1t 1
I = - r 12 Sin
2
(a) da = -'!L
nns 21t J
o
m 2
Peak Inverse Voltage ( PIV ) = V m
RMS value of ripple current
y =
Average value of the current
I = IDe + I'(ripple)
I'(ripple) = (I - IDe)
If =
nns
1 21t )2
- f (I-IOC da
21t 0
= - 2IOC2 + IOC
2
I'nns=
I'rms
y=--
IOC
(
IrmsJ2 -1
IOC
,
I
rms
---
I
OC
1 21t( 2)
- f 12 - 2I.IOC + IOC .da
21t 0
1m X1t
( )
2
for Half Wave Rectifier, y = 2 x 1m -1
P
3.2.7 RATIO OF RECTIFICATION (T'I): DC
PAC
But
DC power delivered to the load
Ratio of Rectification = --. --'------------
AC mput power from transfer secondary
P
De
= Ibex RL
1m
I for HWR=-
De 1t
P = (Im)2 xRL
o. DC n'
P Ae is what a Wattmeter would indicate if placed in the HWR circuit with its voltage terminal
connected across the secondary of the transformer.
From 1t - 21t the diode is not conducting. Hence lac = o.
Pac = (Inns)2 (R
f
+ R
L
)
Q from 1t - 21t the AC is not being converted to D.C. So this power is wasted, as heat across
diode and transformer.
Rectifiers, Filters and Regulators
145
I 2lt I
- fl2 sin
2
ada =-'!!...
Inns( 1t - 21t ) portion of AC is = 21t m 2
It
Inns for HWR is I
m
/2. During negative half cycle, diode is not conducting and current I in
the loop is zero even though Voltage V is present.
1m 1m
( )
2 ( )2
PAC = 2 (Rf+R
L
) = 2 X RL
P
oc
4
- = 212 R = -2 =0.406
PAC 1t m X L 1t
Ratio of rectification for HWR = 0.406
0.406
Considering ideal diode. If we consider R
f
also, the expression I + ( :: )
The AC input power is not converted to D.C. Only part of it is converted to D.C and is
dissipated in the load. The balance of power is also dissipated in the load itself as AC power. We
have to consider the rating of the secondary of the transformer. In ratio of rectification we have
considered only the A.C output as the secondary of the transformer.
3.2.8 TRANSFORMER UTILIZATION FACTOR
D.C power delivered to the load
Transformer Utilization Factor (TUF) = AC t· ft Co d
ra mg 0 ranSiormer secon ary
PAC rated
This term TUF is not ratio of rectification, because all the rated current of the secondary is
not being drawn by the circuit.
Pac = V
nns
x Inns
Vm
V nns = .fi = Rated voltage of (Secondary Transformer)
Inns = Im
/2
Pac =:= V nns x Inns
But V m = 1m (Rr + R
L
) :::: 1m RL ( R
f
is small )
R jlmRL 1m 2..[i
TUF= n
2
x L J2 x"2= 7 =0.287
Transformer Utilization Factor for Half Wave Rectifier is 0.287
146 Electronic Devices and Circuits
3.2.9 DISADVANTAGES OF HALF WAVE RECTIFIER
1. High Ripple Factor (1.21)
3. Low TUF ( 0.287 )
For half wave configuration,
2. Low ratio of rectification (0.406)
4. D.C saturation of transformer.
,-------
I = sin
2
a.da =.!.m..
rms 27t 0 2
Because during negative half cycle the diode will not conduct hence i = 0 in the loop from
7t - 27t. So integration is from 0 - 7t only. Even though V is present, i = 0 for one half cycle.
Therefore, Power is zero ( V x I = P = 0, since, I = 0 )
I It
I = - Jim sina.da = I-cosallt = + 11 =
DC 27t 0 27t 0 27t 7t
Ripple Factor for Half Wave Rectification Irms = 1m 12 = ;
IDe 1m 17t
y= = 1.21
7t
"2 = 1.57
y>1
So the ripple voltage exceeds D.C voltage. Hence HWR is a poor circuit for convertingAC to DC.
3.2.10 POWER SUPPLY SPECIFICATIONS
The input characteristics which must be specified for a power supply are:
1. The required output D.C voltage 2. Regulation
3. Average and peak currents in each diode 4. Peak inverse voltage ( PIV)
5. Ripple factor.
3.3 FULL WAVE RECTIFIER ( FWR )
Since halfwave rectifier circuit has poor ripple factor, for ripple voltage is greater than DC voltage,
it cannot be used. So now analyze a full wave rectifier circuit.
The circuit is as shown in Fig. 3.9.

. .
: : i
I I
! U
l' 0: 1t: 21t
I I
o
A.C.
\.r---IA
I I
. .
Waveforms FWRCircuit
Fig 3.9
Rectifiers, Filters and Regulators
During the +ve half cycle, 0] conducts and the current through O
2
is zero
During the - ve half cycle, O
2
conducts and the current through 0] is zero
A centre tapped transformer is used.
147
The total current i flows through the load resistor RL and the output voltage V 0 is taken
across R
L
.
I 2lt. I
For half wave rectifier, circuit, IDe = -2 flm slOada = --'1l
1t 0 1t
For full wave rectifier, circuit, IDe = Twice that of half wave rectifier circuit
IDe = 2 Im
/1t
1 It
For half wave rectifier circuit, Inns = - f I ~ sin ada = .!.m.
21t 0 2
1 It
For full wave rectifier circuit, Inns = 2 x - f I ~ sin
2
ada
21t 0
Y
m
1=---'-'-'-
m R
f
+ RL
R
f
is the forward resistance of each diode.
A centre tapped transformer is essential to get full wave rectification. So there is a phase shift
of 180
0
, because of centre tapping. So 0] is forward biased during the input cycle of 0 to 1t, O
2
is
forward biased during the period 1t to 21t since the input to O
2
has a phase shift of 180
0
compared to
the input to 0]. So positive half cycle for O
2
starts at a full wave rectifier circuit, while the D.C.
current starts through the load resistance RL is twice that of the Half Wave Rectifier Circuit.
Therefore, for Half Wave Rectifier Circuit, V
1m
I =-
DC 1t
For Full Wave Rectifier Circuit,
21m
IDe = ---;-
Hence, Ripple Factor is improved.
3.3.1 RIPPLE FACTOR
1t
2J2 = 1.11
o
i
o
Fig 3.10 FWR voltage output.
I 2 2. = ~ I r m / -Ioc
2
I'rms V1rms -IDe ' Y Ioe
Therefore, y = 1.21 for HWR, and it is is 0.482 for FWR.
148 Electronic Devices and Circuits
3.3.2 REGULATION FOR FWR
The regulation indicates how D.C voltage varies as a function of load current. The percentage of
regulation is defined as
V
NL
- V
FL
% Regulation = V x 100
FL
The variation of DC voltage as a function ofload current is as follows
V
De
= IDe RL
2:rn . RL [.: Irn = R
f
:
rn
R
L
]
_ 2V
rn
R [ .. I = Vrn ]
- 1t(R
f
+R
L
)' L . rn R
f
+RL
_ 2V
rn
[1- R
f
]
- 1t R
f
+ RL
2V
rn
2V
rn
=--
1t 1t(R
f
+ Rd
I V De = ~ - IDe' Rr I
Vrn
V =-
NL 1t
Vrn
V FL = - - IDe' Rr
1t
2V rn _ 2V rn + Ioc . R
f
% Regulation
~ 1 t ~ __ ~ 1 t ~ _______ x 100
2V
rn
-I
DC
.R
f
1t
I
DC
R
f
= x 100
2V
rn
-I R
dc· f
1t
( 1
l
2V
rn
x (R
f
+Rd -lJ
1t 2Vrn .R
f
1t
R
% Regulation = ~ x 100
L
x 100
Rectifiers, Filters and Regulators
3.3.3 RATIO OF RECTIFICATION
D.C. powerdeliveredto Load
A.C. input power from transformer secondary
PDe = 16e x RL for FWR,
= 21m = 4.1m
2
R
IDe 1t 1[2 V
149
PAC is what a Wattmeter would indicate if placed in the FWRCircuitwith its voltage terminals
connected across the transformer secondary.
1m
Pac = (I
rms
f. (R
F
+R
L
); Irms for FWR = .fi
( ~ )'.(RF + Rd
2
~ ( R +R ).
2 F L
If we assume that R
f
is the forward resistance of the diode is very small, compared to R
L
.
Rr+RL =R
L
·
e
PAC = -I!L(Rd·
2
Poe 4Im2RLx2 8
Ratio of Rectification = --= 2 2 = -2 =0.812
PAC 1t I m x R L 1t
For HWR it is 0.406. Therefore, for FWR the ratio of rectification is twice that of HWR.
3.3.4 TRANSFORMER UTILIZATION FACTOR: TUF
In fullwave rectifier using center-tapped transformer, the secondary current flows through each
half separately in every half cycle. While the primary of transformer carries current continuously.
Hence TUF is calculated for primary and secondary windings separately and then the average
TUF is determined.
DC power del,vered to load
Secondary TUF = ---. ..:.----'------
AC rating of transformer secondary
=
if R
f
« RL ' then secondary TUF = 0.812.
( ~ r · R L
Vm 1m
-x-
J2 J2
150 Electronic Devices and Circuits
The primary of the transformer in feeding two HWR's separately. These two HWR's work
independently of each other but feed a common load. Therefore,
0.287
TUF for primary = 2 x TUF of HWR = 2 x --
I
Rf
+-
RL
ifR
f
« R
L
, then TUF for primary = 0.574
The average TUF for FWR using center tapped transformer
TUF of primary + TUF of secondary 0.812 + 0.574
2 2
= 0.693
Therefore, the transformer is utilized 69.3% in FWR using center tapped transformer.
3.3.5 PEAK INVERSE VOLTAGE (PIV) FOR FULL WAVE RECTIFIER
With reference to the FWR circuit, during the -ve half cycle, D, is not conducting and D2 is
conducting. Hence maximum voltage across RL is V m' since voltage is also present between A
and 0 of the transformer, the total voltage across D, = V m + V m = 2V m ( Fig. 3.9 ).
3.3.6 D.C. SATURATION
In a FWR, the D.C. currents I, and 12 flowing through the diodes D, and D2 are in opposite
direction and hence cancel each other. So there is no problem of D.C. current flowing through the
core of the transformer and causing saturation of the magnetic flux in the core.
PIV is 2 V m for each diode in FWR circuit because, when D, is conducting, the drop across
it is zero. Voltage delivered to RL is V m. D ~ is across R
L
. During the same half cycle, D2 is not
conducting. Therefore, peak voltage across It is V m from the second half of the transformer. The
total voltage across D2 is V m + V m = 2V m·
3.4 BRIDGE RECTIFIERS
The circuit is shown in Fig. 3.11. During the positive half cycle, D
J
and D2 are forward biased.
D3 and D4 are open. So current will flow through D
J
first and then through RL and then through
D2 back to the ground. During the -ve half cycle D 4 and D ~ are forward biased and they conduct.
The current flows from D3 through RL to D
4
. Hence the dIrection of current is the same. So we
get full wave rectified output.
In Bridge rectifier circuit, there is no need for centre tapped transformer. So the transformer
secondary line to line voltage should be one half of that, used for the FWR circuit, employing
two diodes.
II
Fig 3.11 Bridge rectifier circuit.
21m
1t
Rectifiers, Filters and Regulators
where
Rs+2Rf+RL
2V
rn
= ---IDC(RS + 2R
F
)
7t .
Resistance of Transformer Secondary.
= Forward Resistance of the Diode.
= Load Resistance.
151
2R
f
should be used since two diodes in series are conducting at the same time. The ripple
factor and ratio of recti fication are the same as for Fu II Wave Rectifier.
3.4.1 ADVANTAGES OF BRIDGE RECTIFIER
1. The peak inverse voltage (PIV) across each diode is Vm and not 2V
m
as in the
case of FWR. Hence the Voltage rating of the diodes can be less.
2. Centre tapped transformer is not required.
3. There is no D.C. Current flowing through the transformer since there is no
centre tapping and the return path is to the ground. So the transformer utilization
factor is high.
3.4.2 DISADVANTAGES
1. Four diodes are to be used.
2. There is some voltage drop across each diode and so output voltage will be
slightly less compared to FWR. But these factors are minor compared to the
advantages.
Bridge rectifiers are available in a package with all the 4 diodes incorporated in one unit. It
will have two terminals for A.C. Input and two terminals for DC output. Selenium rectifiers are
also available as a package.
3.5 COMPARISON OF RECTIFIER CIRCUITS
HWR FWR Bridge Rectifier
Circuit
Peak Inverse Vo ltage
Vrn
2V
rn
Vrn
Vm 2Vm 2V
rn
No Load Voltage
-- --
1t 1t 1t
Ripple factor 1.21 0.482 0.482
Number of Diodes required 1 2 4
Ratio of Rectification P
dc
0.406 0.812 0.812
PAC
DC Power delivered to the load
TUF=
AC ratings of transformer secondary
0.287 0.693 0.812
152
I
Electronic Devices and Circuits
3.6 VOLTAGE DOUBLER CIRCUIT
3.6.1 HALF WAVE'VOLTAGE DOUBLEwEIRCUIT
The circuit is shown in Fig. 3.12 and 'e wave forms are shown in Fig. 3.13. During the +ve half
cycle, 0
1
is forward biased. So C
1
gets charged to V m. During the -ve half cycle, D
t
jsreverse
biased and the PIV ~ r o s s it V m. Therefore, the total voltage during -ve half cycle V m across C
1
+Vm of -ve half cycle = 2V m. So, O
2
is forward biased during the -ve half cycle and C
2
gets
charged to 2V m. This is the no-load voltage. The actual voltage depends upon the value of RL
connected. In A.C. Voltage measurements, the A.C. input is given to a voltage doubler and filter
circuits. The output of D.C. meter (2V m) is proportional to A.C. Input. The meter calibrated in
terms of the rms value of the A.c. Input.
D2
Fig 3.12 Hal/wave voltage doubler circuit.
3.6.2 FULL WAVE VOLTAGE DOUBLER CIRCUIT
~ t
Fig 3.13 Output waveform.
The circuit is shown in Fig. 3.14. Wave forms are shown in Fig. 3.15. During +ve half cycle, 0
1
conducts C
1
gets charged to V m. During -ve half cycle, C
2
charges through O
2
and to V m. RL is
across the series combination of C
1
and C
2
• Therefore, the total output voltage V 0 = 2V m. Here
the PIV across each diode is only V m and the capacitors are charged directly from the input and
not C
2
will not get charged through C
1
During the time period 0 to 21t we have two cycles.
Current flows from 0 to 1t and 1t to 21t also. So it is FW Voltage Doubler Circuit.
Fig 3.14 Fullwave voltage doubler circuit.
3.7 INDUCTOR FILTER CIRCUITS
FILTERS
i
V
o
I
i
--+t
Fig 3.15 Output waveforms.
A power supply must provide ripple free source of power from an A.C. line But the output of a
rectifier circuit contains ripple components in addition to a D.C. term. It is necessary to include a
Rectifiers, Filters and Regulators 153
filter between the rectifier and the load in order to eliminate these ripple components. Ripple
components are high frequency A.C. Signals in the D.C output of the rectifier. These are not
desirable, so they must be filtered. So filter circuits are used.
Flux linkages per ampere of current L = . The ability of a component to develop induced
voltage when alternating current is flowing through the element is the property of the inductor.
Types of Inductors are :
1. Iron Cored 2. Air Cored
An inductor opposes any change of current in the circuit. ,So any sudden change that might
occur in a circuit without an inductor are smoothed out with the presence of an inductor. In the
case of AC, there is change in the magnitude of current with time.
Inductor is a short circuit for DC and offers some impedance for A.C. (XL = jmL). So it can
be used as a filter. AC voltage is dropped across the inductors, where as D.C. passes through it.
Therefore, A.C is minimized in the output.
Inductor filter is used with FWR circuit. Therefore, HWR gives 121 % ripple and using filter
circuit for such high ripple factor has no meaning. FWR gives 48% ripple and by using filter circuit
we can improve it. According to Fourier Analysis, current, I in HWR Circuit is
. 1m 1m. 21m Cos4rot
I=-+-Smmt-- ---
1t 2 1t 3
21m 4Im 4Im cos4rot
Current in the case of FWR is i ::::: --- --cos 2 rot - -----"=----
- 1t 31t 151t
The reactance offered by inductor to higher order frequencies like 4mt can be neglected. So
the output current 1S
i ::::: 21m - 41m cos2rot
- 1t 31t
the fundamental harmonic m is eliminated.
The circuit for FWR with inductor filter is as shown in Fig. 3.16.
An II '""'-"""" .........
IDU =
Fig 3.16 Inductor filter circuit
One winding of transformer can be used as 'L'. For simplification, diode and choke (inductor)
resistances can be neglected, compared with R
L
. The D.C. component of the current is
Vm
RL
Impedance due to L and RL in Series IZI = + (2mL),2
For second Harmonic, the frequency is 2m.
154 Electronic Devices and Circuits
Therefore, A.C. Component of current 1m = V m I R L
2
+ 4oo
2
L2. So substituting these
values in the expression, for current,
21 41
I = ---1!!... - ---1!!... Cos 200 t
n 3n
by inductor to higher order frequencies like 400t etc., we get,
. 2V 4V
I = _m_ - m cos (2oot - e )
nRL
where e is the angle by which the load current lags behind the voltage and is given by
9=Tan-
12roL
RL .
3.7.1 RIPPLE FACTOR
If
Y = Ir·nn/IO.C. = I' nn/IOC loc = 2V min RL
I - 1m _ 4Vm
rnns - .J2 - + 4oo
2
L2
4vm xnR
Ripple factor = L
+4oo
2
L2 x2Vm
2 1
- --x -;======:=
- 3.J2 4oo2L 2
1+ 2
RL
4oo
2
L2
--2- » 1, then
RL
RLx2
Ripple Factor r =
3-v2 x200L

Fig 3.17 Ripple Voltage V V
Ripple factor = Ji,L
3 200L
.......... ( 3.5 )
Therefore, for higher values ofL, the ripple factor is low. IfRL is large, then also y is high.
Hence inductor filter should be used where the value ofRL is low. Suppose the output wave form
from FWR supply is as shown in Fig. 3.17, then,
V yp _ p is the peak to peak value of the ripple voltage. Suppose
Voc = 300V, and V yp_p is lOY.
10 Vy max 5
then V
R
maximum = '2= 5Y. Therefore, V/ms = .J2 =.J2 = 3.54V
. Vrrms _ 3.54 _ 00118
RIpple Factor y = --- ---. .
Vx 300
% ripple = Ripple factor xl 00 % = 1.18
Rectifiers, Filters and Regulators
3.7.2 REGULATION
v = 2ImRL = 2V
m
DC n n
Im·R
L
= V
m
·
Therefore, V DC is constant irrespective of R
L
. But this is true if L is ideal. In practice
2V
m
V DC = ---IDc-Rf
1t
where Rr is the resistance of diode.
155
An inductor stores magnetic energy when the current flowing through it is greater than the
average value and releases this energy when the current is less than the average value.
Another formula that is used for Inductor Filter for
Rc +RL
Ripple Factor = 0.236 ro .......... ( 3.6 )
L
where RL is load resistance Rc is the series resistance of the inductors. This is the same as the
above formula ~
3 2roL
Problem 3.1
1
~ =0.236.
3",2
A FWR is used to supply power to a 20000 Load. Choke Inductors of 20 H inductance and
capacitors of l6,...f are available. Compute the ripple factor using 1. One Inductor filter 2. One
capacitor filter 3. Single L type filter.
Solution
1. One Inductor Filter :
21m 41m 1 (2roL)
I = ~ - 31t cos (200t - ~ ) where ~ = Tan- R
2V
m
IDC = nRL
I ripple RL 100
V= loe = 3 ~ 2 0 0 L =3x1.414x2 =0.074
2. Capacitor filter :
The ripple voltage for a capacitor filter is of Triangular waveform approximately. The rms
Ed V
nns
value for Triangular wave is r::; or r::; •
2",3 2",3
Ed
Erms = 2J3
156 Electronic Devices and Circuits
Suppose the discharging of the capacitor will cut one for one full half cycle .then the change
last by the capacitor
1
Q=rot = IDe X-.
2f
Q
voltage = -
C
i
--+t
Fig 3.18 Vo of Capacitor filter.
1
= = 0.09
4xJj x50xl6xl0-6 x 2000
3. L Type factor:
Problem 3.2
A diode whose internal resistance is 20 0 is to supply power to a 10000 load from a llOV «rms)
source of supply. Calculate (a) The peak load current. (b) The DC load current (e) AC Load
Current (d) The DC diode voltage. (e) The total input power to the circuit. (1) % regulation from
no load to the given load.
Solution
Since only one diode is being used, it is for HWR Circuit.
I = Vm . = 110.Ji = 152.5mA.
(a) m R
f
+ RL 1020
(b)
(e)
(d)
(e)
(I)
Problem 3.3
- ~ = 152.5 = 48 5 A
IDe - . m .
1t 1t
I = ~ = 152.5 = 76.2mA.
nns 2 2
V De = -Im·R
L
= --48.5 x 1 = --48.5V.
1t
Pi = Inns x(Rf+R
L
) =5.920.
Show that the maximum DC output power P
De
= V
De
IDe in a half wave single phase circuit
occurs when the load resistance equals the diode resistance R
f
.
Rectifiers, Filters and Regulators 157
when RL is very large,
or
Problem 3.4
A 1 rnA meter whose resistance is Ion is calibrated to read rms volts when used in a bridge circuit
with semiconductor diodes. The effective resistance of each element may be considered to be
zero in the forward direction and 00 in the reverse direction. The sinusoidal input voltage is
applied in series with a 5 - KO resistance. What is the full scale reading of this meter?
Solution
21
IDe = ---.!!L for bridge rectifier circuit.
1t
V
I - --1!!...
m - RL'
Vm = .J2vrms
2..[i x Vrms
IDe = 1tR ;
L
RL = 5KO + 100 = (5000 + 10)0 = 5010 0
lmA = 2..[i x Vrms
1txSOIO.
1 x 10-
3
h x 5010 = S.S6V.
Vrms= 2 2
3.8 CAPACITOR FILTER
Here X
f
should be smaller than R
L
. Because, current should pass through C and C should get
chargeo. If C value is very small, Xe will be large and hence current flows through R only
(Fig. 3.19) and no filtering action takes place. During +ve half cycle for a H W R circuit, with C
filter, C gets charged when the diode is conducting and gets discharged (when the diode is not
conducting) through R
L
. When the input voltage e = Em Sin rot is greater than the capacitor
voltage, C gets charged. When the input voltage is less than that of the capacitor voltage, C will
discharged through R
L
. The stored energy in the capacitor maintains the load voltage at a high
value for a long period. The diode conducts only for a short interval of high current. The waveforms
are as shown in Fig. 3.20. Capacitor opposes sudden fluctuations in voltage across it. So the
ripple voltage is minimised.
158
AC
Input
Electronic Devices and Circuits
Fig 3.19 FWR Circuit with C filter.
v
o
t
;,
.1 ••
: '
..
. .
•w :
9 1 9
2
rot ==---.
Fig 3.20 Output of capacitor filter circuit.
So the voltage across RL is more or less constant and hence ripple is reduced, as shown in
the graph, the voltage V 0 is closer to DC wave form. At 9) 'C' gets charged and at 9
2
C starts
discharging.
The point at which diode starts conducting and C gets charged is known as cut in point 9 ..
The point at which the diode stops conducting or the capacitor gets discharged is known as
Cut Out Point 9
2
,
C will not start charging at 1t itself. Therefore, even though the second diode is forward
biased, e
c
is almost =::.Y m' So the first diode conducts from e) to e 2 and the second diode
conducts from 1t + 8) to 1t + 8
2
, Diode conducts only for a short of time when input current is high
and charges C to the peak voltage.
Considering one diode, iB = ie + iR ( Fig. 3.19 )
. _ dV =C dec
Ie - C. dt . d .
. e
c
1=-
R R
t
where e
c
is the voltage to which c gets charged.
Rectifiers, Filters and Regulators 159
But e
c
= V m Sin cot.
Therefore, at that point, diode conducts and C gets charged.
81 = cot
l
·
dec
dt = coY m· Cos cot.
General expression for
. C V V
In
S·
IS = co m cos cot + R m cot
By some mathematical manipulation, ib can be written as
is= -; .jl + co
2
R
2
C
2
Sin(cot + 4»
where ~ = Tan-l(coRC)
At an co instant 8
2
, when the capacitor fully discharges, is = 0
Em S· 8 0
coC V cos 8
2
+ - In 2 = .
m R
or 8
2
= tan-
i
(-nRC)
Thus the value of conduction angle depends upon the values of Rand C, and also the value
of is. Ouring the non conducting interval, the capacitor discharges into the load R supplying load
current.
-ic= i
R
·
de e
The circuit equation is -C x ~ = ~
t
or e
c
=; Ae-
t/RL
At cot = 8
2
e
c
= V m sin 8
2
.
The final expression for E
DC
seems a little complicated. So a simplified expression which is
widely used is V DC = ( V m - V
R
.), where ER is the peak to peak ripple voltage. The expression
2
for Ripple Factor,
V'Ae _ 1t+(8
1
-8
2
)
y = V;;; - 2JjcoRC .
where (8
1
- 8
2
) is the angle of conduction 8
c
. For large coRC ( typical value 100), 8 c = 14.6°.
The larger the value of R, the smaller the y.
3.8.1 RIPPLE FACTOR FOR C FILTER
Initially, the capacitor charges to the peak value V when the diode 0\ is conducting. When the
capacitor voltage is equal to the input voltage, the diode stops conductmg or the current through
the diode 01 is zero. Now the capacitor starts discharging through RL depending upon the time
constant C.R
L
. Therefore, RL value is large. Rate of charging is different from rate of discharging.
When the voltage across the capacitor falls below the input voltage and when the diode 02 is
forward biased, the capacitor will again charge to the peak value Vm and the current through D2
becomes zero when Vc =V
m
. Thus the diodes °
1
and 02 conduct for a very short period 81 to 8
2
and 1t + 8 I to 1t + 8
2
respectively ( Fig. 3.21 ).
160 Electronic Devices and Circuits
V,)
i
\
\
I \
I \
I \
I \
I I I
i \ / D,_ II
I \ I
- - __ J - - - __ - - __________ / _ Conducting \
I \ I It!" I
I \ I I
I I I I
I \ I I •
I \ I \
I \ I I
I \ / I
I \ / \
[ .. ..
11
D,
Conducting

) (l)t
Fig 3.21 Output waveforms of a capacitor filter.
T2 = Time period ofthe discharging of the capacitor
T] = Time period of the charging of the capacitor
The amount of charge lost by the capacitor when it is discharging = Because, IDe
is the average value of the capacitor discharge current. This charge is replaced during a short
interval T] during which the voltage across the capacitor changes by an amount = peak to peak
voltage of the ripple V' rp_p'
Q=Vxc.
Q charge = V p-px C
Q charge = Q Discharge.
V'p_pxC =IDC
xT
2'
or
V' = IDe x T2
p-p C
The output waveform can be assumed to be a triangular wave T2 » T](T] +T
2
) = T/2
when the diode is conducting (0 to n). (T] + T2 corresponds to one half cycle).
TIl
T
- "T-
2 = "2 - 2f" 2 - 2f .
, _'ocx1"T __ 1_ 1
V p-p - C2f . 2 - 2f' for the triangular wave, form factor = Jj'
V'
V'
Ims 2Jj
Voe
IDe = RL
V' rms = 4JjfCR
L
'
Rectifiers, Filters and Regulators
Ripple Factor
v'nns
y = V
DC
= 4.J3fCRL
Therefore, Ripple factor may be decreased by increasing Cor RL or both.
Expression for V
DC
:
or
3.9 LC FILTER
V'p-p
VOC=Vrn- 2
V
DC
Voc= V
rn
- 4fC.R
L
Voc (I+-l-)=Vm
4fCR
L
'.
.. V' P P - IDe I _ V
De
- - 'De-
2fC RL
161
In an inductor filter, ripple decreases with increase in R
L
. In a capacitor filter, ripple increases
with increase in R
L
. A combination of these two into a choke input or L-Section Filter should
then make the ripple independent of load resistance.
If L is small, the capacitor will be charged to V rn' the diodes will be cut off allowing a short
pulse of current. As L is increases, the pulses of current are smoothed and made to flow for a
larger period, but at reduced amplitude. But for a critical value of inductance L
c
' either one diode
or the other will be always conducting, with the result that the input voltage V and I to the filter are
full wave rectified sine waves.
The graph of DC output voltage for LC Filter is as shown in Fig. 3.22.
2V
rn
V
DC
= -1t-
for Full Wave Rectifier Circuit. considering ideal elements,
conduction angle is increased when inductor is placed V R
because there is some drop across L. So C charge
i
to V rn' Therefore, the diode will be forward biased for a
longer period.
The ripple circuit which through L, is not Ie --.
a.Howed to develop much ripple voltage across R
L
, if Xc at Fig 3.22 LC filter characteristic.
rIpple frequency IS small compared to R
L
, Because the
current will pass through C only. Since Xc is small and Capacitor will get charged to a constant
voltage. So V 0 across RL will not vary or ripple will not be there. Since for a properly designed LC
Filter,
Xc «RL
and XL» Xc at (0 = 21t/
XL should be greater than Xc because, all the Af3twukl be or-opped across XL itself so that
AC Voltage across C is nil and hence ripple is low.
162 Electronic Devices and Circuits
3.9.1 RIPPLE FACTOR IN LC FILTER
AC Component of current through L is determined by XL'
X
L
= 2coL
RMS value of ripple current for Full Wave Rectifier with L Filter is,
I 4V
m
I rms == 3refiX
L
= RMS Value of Ripple Current for L Filter
I 2
I rms = ~
3v2XL
2V
m
=--
re
I 2 .J2v
DC
.. I rms = 3.J2XL . V
De
= 3X
L
The ripple voltage in the output is developed by the ripple current flowing through Xe'
I I .J2 Xc V
.. V rms = I rms. Xe = )'X' DC
V'rms
Ripple factor = -V--
DC
.J2.xc
y= 3X
L
1
Xe = ~ ,XL = coL
.J2 1 1
Y = )X2fficX2;L
L
XL = 2coL since ripple is being considered at a frequency, twice the line frequency.
If If=60H I = O.83xl0--6
.... ____ z ...... , _ Y LC
3.9.2 BLEEDER RESISTANCE
For the LC filter, the graph between IDe and V
De
is as shown in Fig. 3.23. For light loads, i.e.,
when the load current is small, the capacitor gets charged to the peak value and so the no load
voltage is 2V Ire. As the load resistance is decreased, IDe increases so the drop across other
elements V iz ~ i o d e s and choke increases and so the average voltage across the capacitor will be
less than the peak value of2V mIre. The out put voltage remains constant beyond a certain point IB
and so the regulation will be good. The voltage Voe remains constant even if IDe increases,
because, the capacitor gets charged every time to a value just below the peak voltage, even though
the drop across diode and choke increases. So for currents above I
B
, the filter acts more like an
inductor filter than C filter and so the regulation is good. Therefore, for LC filters, the load is chosen
such that, the I ~ e ~ lB' The corresponding resistance is known as Bleeder Resistance ( R
B
). So
Bleeder Resistance is the value for which loc ~ IB and good regulation is obtained, and conduction
angle is 180°. When RL = R
B
, the conduction angle = 180°, and the current is continuous. So just
as we have determined the critical inductance Le> when RL = R
B
, the Bleeder Resistance,
IDC = the negative peak of the second harmonic term.
Rectifiers, Filters and Rellulators
i
Fig 3.23 LC filter characteristic.
I = 2V
m
DC 1tRs
4V 1
If peak = ---1!L._.
31t XL
2V
m
4V
m
1
1tRs =J;-'X
L
3.x
L
RB = -2-; RB is Bleeder Resistance.
163
I
DC
should be equal to or greater than I peak, since, XL determines the peak value of t ..
ripple component. If XL is large I' peak can be ~ I
DC
and so the ripple is negligible or pure D.C. is
obtained, where XL = 2mL corresponding to the second harmonic. Therefore, R should be at least
equal to this value ofR
B
, the Bleeder Resistor.
3.9.3 SWINGING CHOKE
The value ofthe inductance in the LC circuits should be > Lc the minimum value so that conduction
angle of the diodes is 180
0
and ripple is reduced. But if the current I
DC
is large, then the inductance
for air cored inductors as L = N ~ I I (flux linkages per ampere of current), as I increases, L decreases
and may become below the critical inductance value Lc' Therefore, Iron cored inductors or
chokes are chosen for filters such that the value of L varies within certain limits and when I is
large, the core saturates, and the inductance value will not be < Lc' Such chokes, whose inductance
varies with current within permissible limits, are called Swinging Cllokes. (In general for inductors
~ I I remains constant so that L is constant for any value of current I ) This can be avoided by
choosing very large value ofL so that, even if current is large, Inductance is large enough> Lc But
this increases the cost of the choke. Therefore, swinging choke are used.
Lc ~ R L / 3m
RL = Load Resistance
Lc = Critical Inductance
m = 21tf
At no load RL = 00. Therefore, L should be 00 which is not possible. Therefore, Bleeder
Resistance of value = 3X
L
/2 is connected in parallel with R
L
, so that even when RL is 00, the
conduction angle is 180
0
for each diode.
164 Electronic Devices and Circuits
The inductance of an iron - cored inductor depends up on the D.C. Current flowing through
it, L is high at low currents and low at high currents. Thus its L varies or swings within certain
limits. This is known as Swinging Choke.
Typical values are
Problem 3.5
L = 30 Henrys at I = 20mA and
L = 4 Henrys at I = 100mA.
Design a full wave rectifier with an LC filter (single section) to provide 9V DC at I OOmA with a
maximum ripple of2%. Line frequency f= 60 HZ.
Solution
. J2 1 1 0.83
RIpple factor y = 3 X 2roC x 2roL = LC Il
0.02 = 0.83 orLC = 0.83 = 41.51l
LC 0.02
Voc _ 9V -900
RL = Ioe - 0.1 - .
RL RL
LC ~ 3ro ~ 1130'
RL
But LC shou'd be 25% larger. :. for f= 60 Hz,the value ofLC should be ~ 900
RL 90
LC ~ 900 ~ 900 = 0.1 Henry.
41.5
IfL=O.IH,then C= 0.1 = 415Ilf. This is high value
IfL = IH, then C = 41.51lf.
If the series resistance of L is assumed to be 50 fl, the drop across L is
IDe x R = 0.1 x 50 = 5Y.
Transformer Rating:
Voe = 9V + 5V = 14V
7t
V m = 2' ( 9 + 5 ) = 22V
22
rms value is .J2 = 15.5V
Therefore, a 15.5 - 0 -15.5 V, 1 OOmA transformer is required. PIV of the diodes is 2V m
Because it is FWR Circuit.
PIV = 44V
So, diodes with 44V, 1 OOmA ratings are required.
Rectifiers, Filters and Regulators 165
Problem 3.6
In a FWR with C filter circuit, Vrp-p = 0.8v and the maximum voltage is 8.8 volts (Fig. 3.24 ).
RL = 1000 and C = 1050 f.lf. Power line frequency = 60 Hz. Determine the Ripple Factor and DC
output voltage from the graph and compare with calculated values.
Solution
V/ p-p = 0.8Y.
0.8
V
nns
' = 2.fj =0.231V
V
DC
= Vm - =8.8- =8.4V.
= V'rms = 0.2 = 0.0238 or 2.38%
y
V
DC
8.4
Theoretical values, y
v
S.svt
x 60 x 100 xl 050f.lf
= 0.057 or 5.75%
..

O.Sv
------
+
---. rot
4tR C
V
DC
= =8.46V
Fig 3.24 For Problem 3.6.
3.10 CLC OR 1t FILTER
In the LC filter or L section filter, there is some voltage drop across L. If this cannot be tolerated
and more D.C. output voltage V Q is desired, p filter or CLC filter is to be used. The ripple factor
will be the same as that of L sectIOn filter, but the regulation will be poor. It can be regarded as a
L section filter with Land C
1
' before which there is a capacitor C,i.e. the capacitor filter. The
input to the L section fhter is tlie output of the capacitor filter C. The output of capacitor filter will
be a triangular wave superimposed over D.C.
Now the output Vo is the voltage across the input capacitor C less by the drop across Lt.
The ripple contained in the output of 'c' filter is reduced by the L section filter LtC
t
.
L}
C
Fig 3.25 CLC or 7r Section Filter
166 Electronic Devices and Circuits
3.10.1 1t SECTION FILTER WITH A RESISTOR REPLACING THE INDUCTOR
Consider the circuit shown here. It is a 1t filter with L replaced by R. If the resistor R is chosen
equal to the r e a c ~ c e ofL (the impedance in 0 should be the same), the ripple remains unchanged.
V DC for a single capacitor filter = (V m - IDd4fC) If you consider C
1
and R, the out put
across C
1
is ( V m - I
DC
/4fC). There is some drop across R. Therefore, the net output is
IDe
V m - 4fC - I
DC
x R
The Ripple Factor for 1t sector is
.J2xc
x
c
I
........... ( 3.7 )
So by this, saving in the cost, weight and space of the choke are made. But this is practical
only for low current power supplies.
Suppose a FWR output current is 100 rnA and a 20 H choke is being used in 1t section filter.
Ifthis curve is to be replaced by a resistor, XL = R. Taking the ripple frequency as 2f= 100 Hz,
XL = coL = 21t(2j).L = 41tjL
= 4 x3.14 x50 x20 = 125600
:: 12 KO
Voltage drop across R = 12,000 x 0.1 = 1200V, which is very large.
R
Fig 3.26 FWR Circuit 1t Section I CRC Filter.
3.10.2 EXPRESSION FOR RIPPLE FACTOR FOR 1t FILTER
The output of the capacitor in the case of a capacitor filter is a Triangular wave. So assuming th,e
output V 0 across C to be a Triangular wave, it can be represented by Fourier series as
V'P - P sin 40)t sin 60)t
V=V
DC
- 1t (Sin2cot- 2 + 3 ... )
IDe
But the ripple voltage peak to peak V' p-p = 2fc' ( for Ll wave)
Neglecting 4th and higher harmonic, the rms voltage ofthe Second Harmonic Ripple is
v'p - P IDe IDe x 2 _ .J2I
De
1t..[i = 2..[i1tfc = 4 • ./21tfc - 41tfc
I
--=x
41tfc c
Rectifiers, Filters and Regulators 167
It is the capacitive reactance corresponding to the second harmonic.
V'rms = fi1ocXc. (across 'C' filter only)
The output ofC filter is the input for the L section filter of LIC
I
. Therefore, as we have
done in the case of L section filter,
. . fiIocXc
Current through the mductor ( harmoOlc component) = _--"'-'''---''''-
XL
I
Output Voltage = Current ( Xc) ( V 0 = I.Xc )
fiIocXc X
V, rms = X . C
I
( after L section)
LI
v'rms _ (fi.loc.xc) XCI
y= -
Voc Voc XLI
Ripple factor,
VOC = IOC·R
L
Y =
RL XLI
where all reactances are calculated at the 2nd harmonic frequency (0 = 2rc/
DC output voltage = ( The DC voltage for a capacitor filter) - ( The drop across LI ).
Problem 3.7
Design a power supply using a rc-filter to give DC output of25V at 100mA with a ripple
factor not to exceed 0.01%. Design of the circuit means, we have to determine L,C, diodes and
transformers.
Fig 3.27 Peak to Peak detector.
Solution
Design of the circuit means, we have to determine L, C, Diodes and Transformer
R = V DC = 25V = 25V = 2500
L Ioc Ioc 100mA
_ Xc _xCI
Ripple factor y - v L.. . .
RL XLI
Xc can be chosen to be = Xc I .
168
X 2
Y = J2. e
RL.x
Ll
·
This gives a relation between C and L.
C2 L =y
There is no unique solution to this.
Electronic Devices and Circuits
Assume a reasonable value of L which is commercially available and determine the
corresponding value of capacitor. Suppose L is chosen as 20 H at 1 OOmA with a D.C. Resistance
of 3750 ( ofInductor).
or
v

Un regulated
D.C.Input
V
1
Regulator
Circuit
Fig. 3.31 Block diagram 01 voltage regulator with D.C input.
D.C.
Output
Voltage
The term Voltage Stabilizer is used, if the output voltage is AC and not DC. The circuits used
for voltage stabilizers are different. The voltage regulator circuits are available in IC form also. Some
of the commonly used ICs are, IlA 723, LM 309, LM 105, CA 3085 A.
7805,7806,7808,7812,7815 : Three terminal positive Voltage Regulators.
7905, 7906, 7908, 7912, 7915 : Three terminal negative Voltage Regulators.
The Voltage Regulator Circuits are used for electronic systems, electronic circuits, IC
circuits, etc.
The specifications and Ideal Values of Voltage Regulators are :
Specifications Ideal Values
l. Regulation (Sv) 0 %
2. Input Resistance (R.) 00 ohms
I
3. Output Resistance (Ro) 0 ohms
4. Temperature Coefficient (ST) 0 mv/oc.
5. Output Voltage V 0
6. Output current range (IL)
7. Ripple Rejection 0 %
Different types of Voltage Regulators are
1. Zener regulator
2. Shunt regulator
3 . .series regulator
4. Negative voltage regulator
5. Voltage regulator with foldback current limiting
6. Switching regulators
7. High Current regulator
Rectifiers, Filters and Regulators 175
3.12.1 ZENER VOLTAGE REGULATOR CIRCUIT
A simple circuit without using any transistor is with a zener diode Voltage Regulator Circuit.
In the reverse characteristic voltage remains constant irrespective of the current that is flowing
through Zener diode. The voltage in the break down region remains constant. (Fig. 3.32)
Fig. 3.32 Zener diode reverse characteristic.
Therefore in this region the zener diode can be used as a voltage regulator. If the output voltage
is taken across the zener, even if the input voltage increases, the output voltage remains constant.
The circuit as shown in Fig. 3.33.
+
Fig. 3.33 Zener regulator circuit
The input Vi is DC. Zener diode is reverse biased.
If the input voltage Vi increases, the current through Rs increases. This extra current flows,
through the zener diode and not through R
L
. Therefore zener diode resistance is much smaller than
RL when it is conducting. Therefore IL remains constant and so Vo remains constant.
The limitations of this circuits are
1. The output voltage remains constant only when the input voltage is sufficiently large so
that the voltage across the zener is V z'
2. There is limit to the maximum current that we can pass through the zener. If Vi is increased
enormously, I
z
increases and hence breakdown will occur.
3. Voltage regulation is maintained only between these limits, the minimum current and the
maximum permissible current through the zener diode. Typical values are from 10m A
to 1 ampere.
3.12.2 SHUNT REGULATOR
The shunt regulator uses a transistor to amplify the zener diode current and thus extending the
Zener's current range by a factor equal to transistor h
FE
• (Fig. 3.34)
176 Electronic Devices and Circuits
Unregulated
D.C.
V
I
B
Fig. 3.34 Shunt Regulator Circuit
Zener current, passes through RI
Nominal output voltage
= V
z
+ V
EB
+
The current that gets branched as IB is amplified by the transistor. Therefore the total current
10 = (P + 1) I
B
, flows through the load resistance R
L
. Therefore for a small current through the
zener, large current flows through ~ L and voltage remains constant. In otherwords, for large current
through RL' Vo remains constant. Voltage Vo does not change with current.
Problem 3.13
For the shunt regulator shown, determine
I. The nominal voltage
2. Value of R
I
,
3. Load current range
4. Maximum transistor power dissipation.
S. The value ofRs and its power dissipation.
Vi = Constant. Zener diode 6.3V, 200mW, requires SmA minimum curre,nt
Transistor Specifications:
V EB = O.2V, hFE = 49, leBo = O.
I. The nominal output voltage is the sum of the transistor V EB and zener voltage.
V 0 = 0.2 + 6.3 = 6.SV = V Eb + V z
2. R( must supply SmA to the zener diode:
8 -6.3 1.7
R = = 340 n
I Sx10-
3
Sx10-
3
3. The maximum allowable zener current is
Power rating 0.2
----=--= - = 318 mA
Voltage rating 6.3 .
The load current range is the difference between minimum and maximum current through the
shunt path provided by the transistor. At junction A, we can write,
IB = I
z
- II
II is constant at S rnA
IB = I
z
-1(
II is constant at 5 rnA
.. IB = S x 10-
3
- S x 10-
3
= 0
Rectifiers, Filters and Regulators
18 (min) = I
Z
Max - 12
= 31.8 x 10-
3
- 5 x 10-
3
= 26.8 rnA
The transistor emitter current IE = 18 + Ie
Ie = P 18 = hFE 18
IE = (P + I) Is = (hFE + I) Is
Is ranges from a minimum of 0 to maximum of 26.8 rnA
Total load current range is (hFE + I) Is
= 50 (26.8 x 10-
3
) = 1.34 A
4. The maximum transistor power dissipation occurs when the current is maximum IE :::: Ie
P D = VolE = 6.5 (1.34) = 8.7 W
5. Rs must pass 1.34 A to supply current to the transistor and R
L
.
V· - Vo 8-6.5
- 1 2
Rs - 1.34 1.34 = 1.1 n
The power dissipated by R
s
'
= IS2 Rs
= (1.34)2 . (1.12) = 3W
REGULATED POWER SlJPPLV
177
An unregulated power supply consists of a transformer, a rectifier, and a filter. For such a circuit
regulation will be very poor i.e. as the load varies (load means load current) [No load means no load
current or 0 current. Full load means full load current or short circuit], we want the output voltage
to remain constant. But this will not be so for unregulated power supply. The short comings of the
circuits are :
I. Poor regulation
2. DC output voltage varies directly as the a.c. input voltage varies
3. In simple rectifiers and filter circuits, the d.c. output voltage varies with temperature also,
if semiconductors devices are used.
An electronic feedback control circuit is used in conjuction with an unregulated power supply
to overcome the above three short comings. Such a system is called a "regulated
power supply".
Stabilization
The output voltage depends upon the following factors in a power supply.
1. Input voltage VI
2. Load current IL
3. Temperature
Change in the output voltage il V 0 can be expressed as
avO avo avo
ilV = --.ilV.+--.ilI
L
+--.ilT
o aV
j
1 oiL aT
ilVo = SI ilVj + Ro illL + Sr ilT
Where the three coefficients are defined as,
178 Electronic Devices and Circuits
(i) Stability factors.
LlV
O
I
Sv = Ll V. Ll IL ==0, LlT ==0
1
This should be as small as possible. Ideally ° since Vo should not change even if Vi changes.
(ii) Output Resistance
LlV
O
I
Ro = LlIL Ll Vi ==0, LlT ==0
(iii) Temperature Coefficient
LlV I
ST == Ll; Ll VI ==0, LlIL ==0
The smaller the values of the three coefficients, the better the circuit.
Series Voltage Regulator
The voltage regulation (i.e., change in the output voltage as load voltage varies (or input voltage
varies) can be improved, if a large part of the increase in input voltage appears across the
control transistor, so that output voltage tries to remain constant, i.e., increase in Vi results in increased
V CE so that output almost remains constant. But when the input increases, there may be some
increase in the output but to a very smaller extent. This increase in output acts to bias the control
transistor. This additional bias causes an increase in collector to emitter voltage which will compensate
for the increased input.
If the change in output were amplified before being applied to the control transistor, better
stabilization would result.
Series Voltage Regulator Circuit is as shown in Fig. 3.35.
r
0
R3
Samphng
Y resistor
I
Control
I1Y=Y
I I
J
R2
Fig. 3.35 Series Regulator Circuit
3.12.3 SERIES VOLTAGE REGULATOR CIRCUIT
RL
r
I1Y
o
=Y
o
L
Q is the series pass element of the series regulator. Q acts as the difference amplifier. D is the
reference zener diode. A fraction of the output voltage b:V 0 (b is a fraction, which is taken across R2
and the potef!.tiometer) is compared with the reference voltage YR. The difference (bVp - V
R
) is
amplified by the transistor Q2. Because the emitter of Q
2
is not at ground potential, there IS constalll
Rectifiers, Filters and Regulators 179
voltage YR. Therefore the net voltage to the Base - Emitter of the transistor Q
2
is (bVo - V
R
). As Vo
increases, (b V Q - V R) increases. When input voltage increases by A Vi' the base-emitter voltage of Q
2
increases. So Collector current of Q
2
increases and hence there will be large current change in R
3
.
Thus all the change in Vi will appear across R
J
!tself. V BE of the transistor Q\ is small. Therefore the
drop across R3 = V CB ofQ\:::: V CE ofQ\ since VeE is small. Hence the increase in the voltage appears
essentially across Q\ only. This type of circuit takes care of the increase in input voltages only. If the
input decreases, output will also decrease. (If the output were to remain constant at a specified value,
even when Vi decreases, buck and boost should be there. The tapping of a transformer should be
changed by a relay when Vi changes). ro is the output resistance of the unregulated power supply
which preceds the regulator circuit. ro is the output resistance of the rectifier, filter circuit or it can
be taken as the resistance of the DC supply in the lab experiment.
AV
The expression for Sv (Stability factor) = A yO
,
= [Rl + R
2
]. (Rl in parallel with R
2
)+ h,e2 +(1 + h
fe2
)R
2
R2 hre2
R
3
R
z
= Zener diode resistance (typical value)
R3 +h
ie1
r
o
+-----"----'-"-'-
1+ h
fe1
R = .
o I+G
m
(R3+ ro)'
The preset pot or trim pot R3 in the circuit :s called as sampling resistor since it controls or
samples the amount of feedback.
Preregulator
It provides constant current to the collector of the DC amplifier and the base of control element.
R3 + h
iel
If R3 is increased, the quantity also increases, but then this term is very small,
1+ hfe
since it is being divided by h
fe
.
3.12.4 NEGATIVE VOLTAGE REGlJLATOR
Sometimes it is required to have negative voltage viz., -6V, -18V, -21 V etc with positive terminal
grounded. This type of circuit supplies regulated negative voltages. The input should also be
negative DC voltage.
3.12.5 VOLTAGE REGlJLATOR WITH FOLDBACK ClJRRENT LIMITING
In high current voltage regulator circuits, constant load current limiting is employed. i.e. The load
current will not increase beyond the set value. But this will not ensure good protection. So foldback
current limiting is employed. When the output is shared, the current will be varying. The series pass
transistors will not be able to dissipate this much power, with the result that it may be damaged. So
when the output is shorted or when the load current exceeds the set value, the current through the
series transistors decrease or folds back.
3.12.6 SWITCHING REGlJLATORS
In the voltage regulator circuit, suppose the output voltage should remain constants at +6V, the input
can be upto + 12V or + 15V maximum. If the input voltage is much higher, the power dissipation
180 Electronic Devices and Circuits
I
across the transistor will be large and so it may be damaged. So to prevent this, the input is limited to
around twice that of the input. But if, higher input fluctuations were to be tolerated, the voltage
regulators IC is used as a switch between the input and the output. The input voltage is not connected
permanently to the regulator circuit but ON/OFF will occur at a high frequency (50 KHz) so that
output is constantly present.
A simple voltage regulator circuit using CA3085A is as shown in Fig. 3.36. It gives 6 V
constant output upto 100 rnA.

2
Un regulated
mput
3
+6Y
Fig. 3.36 CA 3085A IC voltage regulator.
= R2
Rl +R2
RCA (Radio Corporation of America) uses for the ICs, alphabets CA. CA3085A is voltage
regulator. LM309 K - is another Voltage Regulator IC.
I.lA716C is head phone amplifier, delivers 50 mW to, 500-6000 load
CA3007 is low power class AB amplifier, and delivers 30 m W of output power.
MCI554 is 20W class B power amplifier.
Preregulator
The value of the stability factor Sy of a voltage regulator should be very small. Sy can be improved
if R3 is increased (from the general expression) since R3 :::: (VI - V
o
)/!. We can increase R3 by
decreasing I, tllrough R
3
. The current I through R3 can be decreased by using a Darlington pair for
Q,. To get even better values of Sy, R3 is replaced by a constant current source circuit, so that
R3 tends to infinity (R3 00). This constant current source circuit is often called a transistor
preregulator. Vi is the maximum value of input that can be given.
Short Circuit Overload Protection
Overload means 0\ erload current (or short circuit). A power supply must be protected further from
damage through overlo'ad:- In a simple circuit, protection is provided by using a fuse, so that when
current excess of the rated values flows, the fuse wire will blow off, thus protecting the components.
This fuse wire is provided before roo Another method of protecting the circuit is by using diodes.
Rectifiers, Filters and Regulators 181
+
From amplifier 02 of series voltage regulator
Fig. 3.3 7 Circuit for short circuit protection.
Zener diodes can also be employed, but such a circuit is relatively costly.
The diodes DI and D2 will start conducting only when the voltage drop across Rs exceeds the
current in voltage of both the diodes DI and D
2
. In the case of a short circuit the current Is will
increase upto a limiting point determined by
V
Y1
+ V
Y2
- V
BE1
Is = R
S
When the output is short circuited, the collector current of Q
2
will be very high Is . Rs will
also be large.
The two diodes DI and D2 start conducting.
The large collector current of Q
2
passes through the diodes D I and D2 and not through
the transistor QI.
Transistor Q
I
will be safe, DI and D2 will be generally si diodes, since cut in voltage is
0.6V. So Is Rs drop can be large.
Problem 3.14
Design a series regulated power supply to provide a nominal output voltage of 25V and supply load
current IL :s IA. The unregulated power supply has the following specifications Vi = 50 ± 5V,
and ro = 10 n.
Given: R
z
= 12 n at I
z
= 10 rnA. At IC2 = 10 rnA, ~ = t20, h
ie2
= 800 n, h
fe2
= 200, II = 10 rnA.
Vo
The reference diodes are chosen such that V R:::: """"2.
Vo
""""2 = 12.5 V
Two Zener diodes with breakdown voltages of 7.5 V in series may be connected.
R
z
= 12 n at I
z
= 20 rnA
Choose IC2 :::: IE2 = lOrnA
At IC
2
= lOrnA, the h-parameters for the transistor are measured as,
~ = 220, h
ie2
= 800 n, h
fe2
= 200
Choose II = 10 rnA, so that I
z
= IDI + ID2 = 20 rnA
Vo - V
R
25 -15
R = =IKn
D ID 10
182 Electronic Devices and Circuits
IC2 lOrnA
IB2 = -p- = 220 = 451lA
Choose I) as 10 rnA transistors, V
BE
= 0.6 V
V
2
= V
BE2
+ V
R
= 15.6 V
VO-V2 25-15.6
R = = 940 Q
) II lOx 10-
3
V
2
15.6
R
2
:::: -1- = -3 = 1,560 Q
I lOxlO
For the transistor QI' choose 12 as lA and h
FE
) = 125 (d.c. current gain p)
IL +11 +ID
I
B
) = h (A) . Ic:::: IE = IL + I) + ID
fel fJ ) )
(DC Current gain)
1000+ 10+ 10
125 :::: 8 rnA
The current through resistor R3 is I = I
B
) + IC2 = 8 + 10 = 18 rnA
The value of R3 corresponding to Vi = 45 V and IL = IA is (since these are given in the
problem) VI = 50 + 5, 1t5 V
Vi -( YBE
I
+ Vo) 50-25.6
R3 = I = 3 = 1,360 Q.
18x10-
Voltage regulator is a circuit which maintains constant output voltage, irrespective of the changes
of the input voltage or the current.
Stabilizer - If the input is a.c, and output is also a.c, it is a stabilizer circuit.
3.13 TERMINOLOGY
Load Regulation: It is defined as the % change in regulated output voltage for a change in load
current from minimum to the maximum vahre.
E) = Output V<)!tage when IL is minimum (rated value)
E2 = Output voltage when IL is maximum (rated value)
E
I
-E
2
% load regulation = E x 100 %; E) > E
2
. This value should be small.
1
Line Regulation : It is the % change in Vo for a change in Vi.
flVo
= -- x 100 %
flV
j
In the Ideal case fl V 0 = 0 when Vo remains constant.
This value should be minimum.
Load regulation is with respect to change in IL
Line regulation is with respect to change in V ..
Ripple Rejection : It is the ratio of peak to peak output ripple voltage to the peak to peak input
ripple voltage.

V;(P-p)
Rectifiers, Filters and Regulators 183
SUMMARY
• Rectifier circuit converts AC to Unidirectional Flow.
• Filter Circuit converts Unidirectional flow to DC. It minimises Ripple.
• The different types of filter circuits are
1. Capacitor Filter
2. Inductor Filter
3. L-Section ( LC ) Filter.
4. 1t-Section Filter
5. CRC and CLC Filters
I I
• Ripple Factor = For Half Wave Rectifier, y = 1.21, for Full Wave Rectifier
IDe
y = 0.482
• Expression for Ripple Factor for C Filter,
1
y= 4.fjjCR
L
• ExpressiollJor Ripple Factor for L Filter,
RL
y = 4.fjroL
• Expression for Ripple Factor for LC Filter,
.J2 1 1
y= 3
x
2roC x 2roL
• Expression for Ripple Factor for 1t Filter,
y=
RL XLI
R
• Critical Inductance Lc
3X
L
• Bleeder Resistance RB = -2-
OBJECTIVE TYPE QUESTIONS
I. Rectifier Converts .............................. to ............................ ..
2 .. Filter Circuit Converts .............................. to ............................ ..
3. Ripple Factor in the case of Full Wave Rectifier Circuit is ............................ ..
4. The characteristics of a swinging choke is ............................. .
5. In the case of LC Filter Circuits, Bleeder Resistance ensures
(1) ............................ ..
(2) ............................. .
184 Electronic Devices and Circuits
ESSAY TYPE QUESTIONS
1. Obtain the expression for ripple factor in the case of Full Wave Rectifier Circuit with
Capacitor Filter.
2. Explain the terms Swinging Choke and Bleeder Resistor.
3. Compare C, L, L-Section, x-Section (CLC and CRC ) Filters in all respects.
MULTIPLE CHOICE QUESTIONS
1. Special types of diodes in which transition time and storage time are made small
are called ...
(a) Snap diodes (b) Rectifier diodes (c) Storage d iodes (d) Memol)' diodes
2. The Circuit which converts undirectional flow to D.C. is called
(a) Rectifier circuit (b) Converter circuit
(c) filter circuit (d) Eliminator
3. For ideal Rectifier and filter circuits, % regulations must be ...
(a) 1% (b) 0.1 % (c) 5% (d) 0%
4. The value of current that flows through RL in a 'n' section filter circuit at no load is
(a) 00 (b) 0.1 rnA (c) 0 (d) few rnA
•
,cs
In tltis Chapter,
• The principle of working of Bipolar Junctions Transistors, (which are also
simply referred as Transistors) and their characteristics are explained.
• The Operation of Transistor in the three configurations, namely Common
Emitter, Common Base and Common Collector Configurations is explained.
• The variation of current with voltage, in the three configurations is given.
• The structure of Junction Field Effect Transistor (JFET) and its V-I
characteristics are explained.
• The structure of Metal Oxide Semiconductor Field Effect Transistor
(MOSFET) and its V-I characteristics are explained.
• JFET amplifier circuits in Common Source (CS) Common Drain (CD) and
Common Gate (CG) configurations are given.
186 Electronic Devices and Circuits
4.1 BIPOLAR JUNCTION TRANSISTORS ( BJT'S )
The device in which conduction takes place due to two types of carriers, electrons and holes is
called a Bipolar Device. As p-n junctions exist in the construction of the device, it is a junction
device. When there is transfer of resistance from input side which is Forward Biased ( low
resistance) to output side which is Reverse Biased ( high resistance ), it is a Trans Resistor or
Transistor Device. There are two types of transistors NPN and PNP. In NPN Transistor, a
p-type Silicon (Si) or Germanium (Ge) is sandwiched between two layers of n-type silicon. The
symbol for PNPtransistor is as shown in Fig. 4.1( a) and forNPN transistor, is as shown in Fig4.1 (b).
E
(a)
PNP
C
E
(b)
NPN
Fig. 4.1 Transistor symbols.
C
The three sections of a transistor are Emitter, Base and Collector. If the arrow mark is
towards the base, it is PNP transistor. If it is away then it is NPN transistor. The arrow mark on
the emitter specifies the direction of current when the emitter base junction is forward biased.
When the PNP transistor is forward biased, holes are injected into the base. So the holes move
from emitter to base. The conventional current flows in the same direction as holes. So arrow
mark is towards the base for PNP transistor. Similarly for NPN transistor, it is away. DC Emitter
Current is represented as IE' Base Current as IB and Collector Current as Ie' These currents are
assumed to be positive when the currents flow into the transistor. V EB reters to Emitter - Base
Voltage. Emitter ( E ) Voltage being measured with reference to base B ( Fig. 4.2). Similarly
V CB and VCE'
-+IE
PNP
+E Ic -+-- C+
t t
V
EB
VCB
~
IB t
~
----------
---------_.
B
Fig. 4.2 Current components in a pnp transistor.
4.1.1 POTENTIAL DISTRIBUTIONS THROllGH A TRANSISTOR
Fig. 4.3 shows a circuit for a PNP transistor. Emitter Base Junction is Forward Biased Collector
- Base Junction is reverse biased. Fig. 4.4 ( a) shows potential distribution along the transistor (x).
Transistor Characteristics 187
When the transistor is open circuited, there is no voltage applied. The potential barriers adjust
themselves to a height V . V will be few tenths of a volt. It is the barrier potential. So when the
transistor is open circuitt?d h01es will not be injected into the collector. (Potential barrier exists all
along the base width, since the holes from emitter have to reach collector and not just remain in base).
Fig 4.3 PNP transistor circuit connections.
But when transistor bias voltages are applied, emitter base junction is forward biased and
the collector base junction is reverse biased. The base potential almost remains constant. But
if the reverse bias voltage of collector - base junction is increased, effective base width decreases.
Since EB junction is forward biased, in a PNP transistor, the emitter base barrier is lowered. So
holes will be injected into the base. The injected holes diffuse into the collector across the n-type
material (base). When EB junction is forward biased emitter - base potential is increased by IV EBI.
Similarly collector base potential is reduced by IV cBI. Collector base junction is reverse biaseo.
So collector extends into the base or depletion region width increases. Emitter - Base Junction is
to be Forward Biased because, the carriers must be injected and Collector - Base Junction must
be reverse biased, because the carriers must be attracted into the Collector Region. Then only
current flow results through the transistor.
4.1.2 TRANSISTOR CURRENT COMPONENTS
Consider a PNP transistor. When Emitter - Base junction is forward biased, holes are injected into
the base. So this current is I (holes crossing form Emitter to base). Also electrops can be
injected from base to emitterPr which also contribute for emitter current IE). So the resulting
current is I
nE
. Therefore the total emitter current IE = IpE + I
nE
.
I
The ratio of ~ is proportional to the ratio of conductivity of the 'p' material to that of'n'
~ .
material. In commercial transistor the Doping of the Emitter is made much large than the
Doping of the Base. So 'p' region conductivity will be much larger than 'n' region conductivity.
So the emitter current mainly consists of holes only. (Such a condition is desired since electron
hole recombination can take place at the emitter junction). Collector is lightly doped. Basewidth
is narrow ( Fig. 4.4(a».
All the holes injected from emitter to base will not reach collector, since some of them
recombine with electrons in the base and disappear.
188 Electronic Devices and Circuits
E B c
(a)
____
(b) tmiuer
: -+-Distance x
V
EB
,
---f -------! - - - -
: Base'
: Width : \L-___ _
, ,
Fig 4.4 Potential barrier levels in a transistor.
If Ip is the current due to holes at the Emitter Junction, IE> I e since some holes have
recombinea with electrons in the base. The recombination curreht is equal to ( IrE -I
pe
)' If the
emitter is open circuited, so that I = 0, then I would be zero. Under these condItions, junction
acts as a reverse biased diode. So tfie Col1ectotCurrent is approximately equal to reverse saturation
Current leo' If IE -:j:. 0, then Ie = leo + Ipc
For a PNP transistor, leo consists of holes moving from left to right (base to col1ector) and
electrons crossing in the opposite direction. In the base, which is n-type, holes ar:e the minority
carriers. In the collector, which is 'p' type, electrons are the minority carriers. I is due to
minority carriers. Currents entering the transistor are positive. For NPN transistor, of
electrons, the minority carriers in the base moving to collector and holes from collector moving to
base. So the direction of leo for NPN transistor is the same as the conventional current entering
into the transistor. Hence leo is positive for NPN transistor.
Emitter Efficiency (y ) :
Current due to Injected at the Emitter - Base Junction
y =
Total Emitter Current
For a PNP transistor,
IpE = IpE
IpE + InE IE
y
Value ofy is always less than I.
IpE = Current due to injected holes from emitter to base
InE = Current due to injected electrons from base to emitter.
Transistor Characteristics
Transportation Factor ( 13 *) :
13* = Current due to injected carriers reaching B - C junction
Injected carrier current at emitter - base junction
For a PNP transistor,
* _ Ipc Ipin the Collector _
13 --= -
I pE I pin the Emitter
Large Signal Current Gain (a )
Collector Current
a=
Emitter Current
Ipc
a= -
IE
Multiplying and dividing by I
pE
'
Ipc IpE
a= - x-
IpE IE
Ipc *
- =13
IpE
4.1.3 TRANSISTOR As AN AMPLIFIER
Hole Current in the Collector
Hole Current in the Emitter
189
A small change V of Emitter - Base Voltage causes relatively large Emitter - Current Change
a' is defined the ratio of the change in collector current to the change in emitter current.
Because of change in emitter current and consequent change in collector current, there will
be change in output voltage V .
o
V 0 = a' x RL X
r' x
I e
RL is load resistance. Therefore, the voltage amplification is,
A = =
v re'ME
a'.R
L

A = --'a' =--
V VCB=K
R > r'
L e
Ay> I

Since a' = --

a' = Small Signal Forward - Circuit Current Gain.
r' = Emitter Junction Resistance.
e
190 Electronic Devices and Circuits
The term, r' is used since this resistance does not include contact or lead resistances.
r' value is small, beecause emitter - base Junction is forward biased.
e
So a small change in emitter base voltage produces large change in collector base voltage.
Hence, the transistor acts as an amplifier. (This is for common base configuration. Voltages are
measured with respect to base ).
The input resistance of the circuit is low (typical value 400) and output resistance is high
(3,0000). So current from low resistance input circuit is transferred to high resistance output
circuit ( l\). So it is a transfer resistor device and abbreviated as transistor. The magnitude
of current remains the same ( a' ~ I ).
4.2 TRANSISTOR CONSTRUCTION
Two commonly employed methods for the fabrication of diodes, transistors and other semiconductor
devices are :
1. Grown type 2. Alloy type
4.2.1 GROWN TYPE
It is made by drawing a single crystal from a melt of Silicon or Germanium whose impurity
concentration is changed during the crystal drawing operation.
4.2.2 ALLOY TYPE
A thin wafer of n-type Germanium is taken. Two small dots of indium are attached to the wafer
on both sides. The temperature is raised to a high value where indium melts, dissolves Germanium
beneath it and forms a saturator solution. On cooling, indium crystallizes and changes Germanium
to p-type. So a PNP transistor formed. The doping concentration depends upon the amount of
indium placed and diffusion length on the temperature raised.
4.2.3 MANlJFACnrRE OF GROWN JlJNCTION TRANSISTOR
Grown Junction Transistors are manufactured through growing a single crystal which is slowly pulled
from the melt in the crystal growing furnace Fig. 4.5. The purified polycrystalline semiconductor is
kept in the chamber and heated in an atmosphere ofN and H2 to prevent oxidation. A seed crystal
attached to the vertical shaft which is slowly pulled at tFte rate of (I mm / 4hrs) or even less. The seed
crystal initially makes contact with the molten semiconductors. The crystal starts growing as the
seed crystal is pulled. To get n-type impurities, to the molten polycrystalline solution, impurities are
,!,
t
1. Furnace
4 5
2. Molten Germanium
3. Seed Crystal
4. Gas Inlet
1
5. Gas Outlet
6. Vertical Drive
Fig 4.5 Crystal growth.
Transistor Characteristics 191
added. Now the crystal that grows is n-type Then p-type impurities are added. So
the crystal now is p-type. Like this NPN transistor can be fabricated.
4.2.4 DIFFlJSION TYPE OF TRANSISTOR CONSTRUCTION
To fabricate NPN transistor, !I-type 'Si' ( or 'Ge' ) wafer is taken. This forms the collector.
Base-collector junction areas is determined by a mask and diffusion length. p-type impurity (Boron)
is diffused into the wafer by diffusion process. The wafer to be diffused and the Boron wafer
which acts as p-type impurity, are placed side by side. Nitrogen gas will be flowing at a particular
rate, before the Boron is converted to Boron Oxide and this gets deposited over Siliconwafer.
Baron is driven inside by drive in diffusion process. So the base-collector junction will be formed.
Now again n-type impurity ( Phosphorus) is diffused into the Base Area in a similar way.
The different steps in process are : B E
1. Oxidation
2. Photolithography
3. Diffusion
4. Metallization
5. Encapsulation
4.2.5 EPITAXIAL TYPE
In this process, a very thick, high purity (high resistance), single crystal
layer of Silicon or Germanium is grown as a heavily doped substrate of
the same material. (n on n +) or (p on p +). This forms the collector.
Over this base and emitter are diffused. Epitaxi is a greek word. Epi
mean 'ON' Taxi means 'ARRANGEMENT'. This technique implies
the growth of a crystal on a surface, with properties identical to that of
the surface. By this technique abrupt step type p-n junction can be
formed. The structure is shown in Fig 4.6 (a).
4.3 THE EBERS - MOLL EQUATION
EBERS - MOLL MODEL OF A TRANSISTOR
Lo:J
p
n - epitai
n - substrate
Fig 4.6 (a) Epitaxial
type transistor.
A transistor can be represented by two diodes connected back to back. Such a circuit for a PNP
transistor shown is called Eber - Moll Model. The transistor is represented by the diodes connected
back to back. lEO and leo represent the reverse saturation currents. aN IE is the current source.
a
I
Ie represents the current source in inverted mode ( Fig. 4.7 ).
Ie aN IE
N ,.--+
p
C
+ +
V
E
B Ve
Fig 4.7 Eber-Moll Model of PNP transistor.
A transistor cannot be formed by simply connecting two diodes since, the base region should
be narrow. Otherwise the majority carriers emitted from the emitter will recombine in the base
and disappear. So transistor action will not be seen.
192 Electronic Devices and Circuits
Similarly
-AeDpPno
a
l2
=
w
1
= (VE/VT_l)+ (vc/vT_l)
c a
21
e a
22
e
-AeDpPno (DpPno DnnE)
a = . . a =Ae --+--
21 w '22 W Lc
.......... (1)
.......... (2)
.. all = a
21
; Equations (1) and (2) are called Eber-Moll Equations.
4.4 TYPES OF TRANSISTOR CONFIGURATIONS
Transistors ( BJTs ) are operated in three configuration namely,
1. Common Base Configuration (C.B)
2. Common Emitter Configuration (C.E)
3. Common Collector Configuration (C.C)
Fig. 4.8 Common base amplifier circuit.
4.4.1 COMMON BASE CONFIGURATION (C.B)
The base is at ground potential. So this is known as Common Base Configuration or Grounded
Base Configuration. Emitter and Collector Voltages are measured with respect to the base. The
convention is currents, entering the transistor are taken as positive and those leaving the transistor
as negative. For a PNP transistor, holes are the majority carriers. Emitter Base Junction is
Transistor Characteristics 193
Forward Biased. Since holes are entering the base or IE is positive. Holes through the base reach
collector. Ie is flowing out of transistor. Hence Ic IS negative; similarly IB is negative. The
characteristics of the transistor can be described as,
V EB = f\(V CB' IE)
Dependent Variables are Input Voltage and Output Current.
Ic = f2 (V CB' IE)
Independent Variables are Input Current and Output Voltage. For a Forward Biased Junction
V EB is positive, for reverse bias collector junction V CB is negative.
Transistor Characteristics in Common Base Configuration
The three regions in the output characteristics of a transistor are
1. Active region
2. Saturation region
3. Cut off region
The input and output characteristics for a transistor in C.B configuration are as shown in
Fig. 4.9 (a) and (b) When IE = 0, the emitter is open circuited. So the transistor is not conducting
and the IfBu!S not considered. Generally the transistor is not used in this mode and it is regarded
as cut of ; Even though ICBO is present it is very small for operation. During the region OA, for
a small variation of V BC there is very large, change in current Ic with IE' So current is independent
of voltage. There is no control over the current. Hence the transistor can not be operated in this
region. So it is named as Saturation region. The literal meaning of saturation should not be taken.
The region to the right is called active region. For a given V cs.' with IE increased, there is no
appreciable change It is in saturation. When transistor IS being used as a switch, it is
operated between cut ott and saturation regions.
IE (mA)
SmA
V = 0.6
T
(a)
Saturation
IV
region
IE = 40 mA
:--J'----=-
I
E
= 30 mA

VCE(V)
Fig 4.13 Output characteristics in C.E Configuration.
196
1+1 +1 =0
BeE
IE = - (IB + Ie)
- Ie = +Ieo - a . IE
But I = - (I + I )
E B e
Ie = - leo + a . IB + a . Ie
Ie (1 - a) = - leo + a . IB
Electronic Devices and Circuits
If the transistor were to be at cut off, IB must be equal to zero.
TEST FOR SATURATION
1.
2.
I
I I> .!s..
B -
eB is for PNP transistor and negative for NPN transistor, the transistor
IS In saturatIOn.
COMMON EMITTER CUT OFF REGION
If a transistor was to be at cut off, Ie = O. To achieve this, if the emitter-base junction is open
circuited, there cannot be injection of carriers from E to C, through B. Hence Ie = O. (Any
reverse saturation current flowing because of thermal agitation is very small. So the transistor can
not be operated). But in the case ofC.E configuration, even ifIB is made 0, the transistor is not cut
off. Ie will have a considerable value even when IB = O. The circuit is shown in Fig. 4.14.
Because,
But
If
or
or
RL
CE
Fig 4.14 Common Emitter Cut-off region.
Ie = - a IE + leo'
a = Current gain ;
leo = Reverse saturation current in the collector.
IE = -(I
B
+ Ie)'
IB = 0, IE = -Ie
Ie = - a(-I
e
) + leo
Ie (1- a) = leo
leo
Ie = 1 - a = leBO
Transistor Ch aracteristics 197
In order to achieve cut-off in the Common Emitter Configuration, Emitter Base Junction
should be reverse biased. a is constant for a given transistor. So the actual collector current with
collector junction reverse biased and base open circuited is designated by !cOB' If a ~ 0.9, then
leBo ~ 10 leo' So the transistor is not cut off. Therefore, in the C.E contlguration to cut of the
transistor, it IS not sufficient, if IB = O. The Emitter Base Junction must be reverse biased, so that
IE is equl to O. Still If. will not be zero, but will be leo' Since Ico is small, the transistor can be
regarded as at cut of .
REVERSE COLLECTOR SATURATION CURRENT leBO
ICBO is due to leakage current flowing not through the j unction, but around it. The collector current
in a transistor, when the emitter current is zero, is designated by IC.Bo· IlcBol can be> leo' leo is
collector reverse saturation current (when collector-base junction IS reverse biased).
SATURATION RESISTANCE
For a transistor operating in the saturation region, saturation resistance for Common Emitter
configuration is of importance. It is denoted by R
eEs
or RcE's (saturation) or Rcs
VCE (sat)
Res = Ic
The operating point has to be specified for Rcs'
Saturation Voltage:
Manufacturers specify this for a given transistor. This is done in number of ways. Res value may
be given for different values of I or they may supply the output and input characteristic for the
transistor itself. V CE(Sat) depen9s upon the operating point and also the semiconductor material,
and on the type of transistor construction. Alloy junction and epitaxial transistors give the lowest
values of V CEO Grown junction transistor yield the highest V CEO
DC CURRENT GAIN J3 de
J3 de is also designated as hFE (DC forward current transfer ratio).
J3 = .!£
IB
Vee
1=-
c RL
So ifJ3 is known, IB = (Ip )
Therefore, .!£ gives the value of base current to operate the transistor in saturation region.
~
J3 varies with Ic for a given transistor.
TEST FOR SATURATION
To know whether a transistor is in saturation or not,
1.
Ie
198
2.
But, .
Electronic Devices and Circuits
V CB should be positive for PNP transistor and negative for NPN transistor. In
the C.E configuration,
Ic = - (IB + IE)
Ic = - alE + Ico
leo - Ie
1=--"'-"'--"'-
E a
- leBo a.l B
1=--+--
c I-a I-a
a
13=-
I-a
a
(1- a) = -

If we replace Ico by ICBo'

Ic = ICBO x ( 13 + I ) + -a-
I Ic = (1 + (3)1CBO + 13 IB I
Transistor is cut off when Ie = 0, Ie = leBO and IB = - ICBO' So at cut-off 13 = O.
THE RELATIONSHIP BETWEEN a AND 13
IE = Emitter Current
IB = Base Current
Ic = Collector Current
Since in a PNP transistor, IE flows into the transistor. IB and Ic flow out of the transistor.
The convention is, currents leaving the transistor are taken as negative ..
IE = - (IB + Ie)
or IE + IB + Ie = 0
or I = - I - I
e E B
-Ie
But a=-
IE
or for PN P transistor taking the convention,
-Ie
1=-
E a
Ie
I = + - - I
CaB
I =-1
e a) B
(a-I)
le-a- =-IB
Transistor Characteristics
or
I (I-a) =+1
CaB
~ - ~
'B I-a
,
f = ~ = Large Signal Current Gain = hFE
B
The Expression for Collector Current, Ie :
199
The current flowing in the circuit, when E - B junctton IS left open i.e., IB = 0 and collector is
reverse biased. The magnitude of I
CEO
is large. Hence the transistor is not considered to be cut
off.
The general expression for Ic in the active region is given by,
Ic = -a IE + I
CBO
IE = - (Ic + I
B
)
Substitute eq. (2) in eq. (I)
where
Ic = + a (Ic + I
B
) + ICBO
Ic [1 - a] = alB + I
CBO
a ,
I = -- I + eBO
C I-a B I-a
Ic = ~ IB + ( ~ + 1) I
CBO
Ic = ~ IB + ICEO
leBo
I =-
CEO I-a
I
CEO
is the reverse collector to emitter current when base is open.
RELATIONSHIP BETWEEN I
CBO
AND I
CEO
Ic = - alE + I
CBO
This is.the general expression for Ic in term of a IE and ICBO·
when IB = 0, in Common-Emitter (C.E. ) Configuration,
Ic = - IE = ICEO
Ic + IB + IE = 0
18 = 0 ( Since E-B Junction is left open)
I =-1
c E
I
CEO
= +a I
CEO
+ ICBO
or ICEO ( 1 - a ) = ICBO
ICBO 1
ICEO = (I - a); (1 - a) = ( ~ + 1)
ICEO = ( ~ + 1) I
CBO
..... ( 1)
..... (2)
200 Electronic Devices and Circuits
In Common Emitter Configuration, for NPN transistor, holes in the collector and electrons in
the base are the minority carriers. When one hole from the collector is injected into the base, to
neutralize this, the emitter has to inject many more electrons. These excess electrons which do
not combine with the hole, travel into the collector resulting in large current. Hence I
CEO
is large
to I
CBO
' When E B junction is reverse biased, the emitter junction is reduced and hence
lCEO IS reduced.
Therefore, the expression for [3 gives the ratio of change in collector current (Ic - I
CBO
) to
the increase in base current from cut off value ICBO to the lB' So [3 is large signal current gain
of a Transistor in Common Emitter ConfiguratIOn.
4.4.3 THE COMMON COLLECTOR CONFIGURATION (C.C)
Here the load resistor RL is connected in the emitter circuit and not in the collector circuit. Input
is given between base and ground. The drop across RL itself acts as the bias for emitter base
junction. The operation of the circuit similar to that of Common Emitter Configuration. When the
base current is I 0' emitter current will be zero. So no current flows through the load. Base
current IB shouldl5e increased so that emitter current is some finite value and the transistor comes
out of cut-off region.
Input characteristics IB vs V BC Output characteristics Ic vs V EC
The circuit diagram is shown in Fig. 4.15 (a) and the characteristics are shown in Fig. 4.15
(b) and (c).
(a) Cirucuitfor C.C configuration
10 rnA
6V VBC(V)
(b) Input characteristics in C.C configuration
IB = 0 J.LA
VEC(V)
(c) Output characteristics in C.C configuration
Fig. 4.15
Transistor Characteristics
The current gain in CC, 'y' :
It is defined as emitter current to the base current
IE
y- -
- - Ia
Relationship between 'a' and 'y' :
Ie
0.- -
- - Ia
IE + IB + Ic = 0, :. IB = (lc + IE)
Substitute eq. (3) in eq. (I)

. ---p - type Gate
Fig 4.25 Structure of JFET.
Drain
The drain D is the terminal through which the majority carriers leave the bar. Current entering
the bar at D is designated by I
D
. V DS is positive if D is more positive than S, source is forward biased.
Gate
In the case of n - channel FET, on both sides of the bar heavily doped (p+) region of acceptor
impurities have been formed by diffusion or other techniques, The impurity regions form as
Gate. Both these region (p+ regions) are joined and a lead is taken out, which is called as the
gate lead of the FET. Between the gate and source, a voltage V GS is applied so as to reverse
bias the gate source p-n junction. Because of this reason JFET has high input impedance.
In BJT Emitter Base junction is forward biased. So it has less Current entering the bar
at G is designated as I
G
.
CHANNEL
The region between the two gate regions is the channel through which the majority carriers move
from source to drain. By controlling the reverse bias voltage applied to the gate source junction
the channel width and hence the current can be controlled.
Transistor Ch aracteristics
4.7 FET STRUCTURE
FET will have gate junction on both sides of the silicon bar (as shown in Fig. 4.26).
Source ~
S ..-.
"(,(,
+ -
+
Fig 4.26 JFET structure.
215
But it is difficult to diffuse impurities from both sides ofthe wafer. So the normal structure
that is adopted is, a p-type material is taken as substrate. Then n-type channel is epitaxially
grown. A p-type gate is then diffused into the n-type channel. The region between the diffused
p-type impurity and the source acts as the drain.
2b (x) = Actual channel width, at any
point x, the spacing between
the depletion regions at any
point x from source end
because the spacing between
the depletion regions is not
uniform, it is less towards the
drain and more towards the
source (see Fig.4.27). 2b
depends upon the value
ofV
Gs
'
2a = Channel width, the spacing
Source
Fig 4.27 Structure
between the doped regions of the gate from both sides. This is the channel width
when V GS = 0 (and the maximum value of channel width).
4.7.1 THE ON RESISTANCE r
DS
(ON)
Suppose a small voltage Vos is applied between drain and source, the resulting small drain current
10 will not have much effect on the channel profile. So the effective channel cross section A
can be assumed to be constant, throughout its. length.
Where,
A = 2bw
2b = Channel width corresponding to zero 10 (Distance between the space
charge regions)
w = Channel dimension perpendicular to b.
216 ElectJ'OJH(: n·;·· .. ices and Circuits
.. Expressions for,
ID = A . e . N D Ex
J = n e. . E (J = cr.E. U .=
I = A x J,
But A = 2b (x) . w
Considering b along x-axis, since 'b' dependf, upon the voltage V GS' b(x) is till' width
at any point 'x'. J
. . ID = 2b (x) roo q. ND . .':,
V .
en = ; L is the length of the channel
(Electric field strength depends upon V DS' Channel width 2b depends upon Vas.)
.. ID = 2b (x) w. q. ND . . Vos
L
'This expression is valid in the linear region only. This is the linear region, It behaves like
a resistance, where ohmic resistance depends upon V as.v DS / ID is called as the ON drain
resistance or rDS(ON) ON because, JFET is conducting, in the ohmic regions when V DS is small
and channel area A = 2a.w,
rDS(on) = Ll2 b(x) wq ND
rDS(on) will be few n to several 100 n.
Because, the mobility of electrons is much higher than that for holes, rDS(On) is small for
n-channel FETs compared to p-channel FETs.
4.7.2 PINCH OFF REGION
The voltage V DS at which, the drain current ID tends to level off is called the pinch off voltage.
When the value of V DS is large, the electric field that appears along the x-axis, i.e., along the
channel ex will also be more. When the value of ID is more, the drain end of the gate i.e.,
the gate region near the drain is more reverse biased than the source end. Because, the drain
is at reverse potential, for n-channel FET, source is n-type, drain is n-type and positive voltage
is applied for the drain, gate is also reverse biased. Therefore, the drop across the channel adds
to the reverse bias voltage of the gate. Therefore, channel narrows more near the drain region
and less near the source region. So, the boundaries of the depletion region are not parallel when
V DS is large and hence Ex is large. .
As V DS increases, Ex and ID increase, whereas the channel width b(x) narrows
(.: depletion region increases). Therefore the current density J = ID / 2b(x) w increases.
If complete pinch off were to take place b = O. But this cannot happen, because if b were
to be 0, J 00 which cannot physically happen.
Mobility is a function of electric field intensity remains constant for low electric fields
when Ex < 103V/cm.
a k when Ex is 10
3
to 10
4
V/cm.(for moderate fields).
u a _1 for very high field strength > 10-1 V!ern.
Ex
'0 = 2bwe ND Ex'
Transistor Characteristics 217
As V DS increases, Ex increases but )..tn decreases, b almost remains constant. Therefore
If) remains constant in the pinch off region.
Pinch off voltage (V PO or V p) is the voltage at which the drain current ID levels off.
When V GS = OV, if V GS = - 1, V, the voltage at which the current ID levels off decreases,
(This is not pinch off voltage). If V GS is large - 5, V DS at which the current levels off may
be zero. So the relationsh ip between V DS' V PD and V GS is
V DS = V PD + V GS V DS is positive n-channel FET.
V PO = P indicates pinch off voltage, zero indicate the voltage when V GS = Ov.
V DS is the voltage at which the current ID levels off.(See Fig. 4.28).
lOrnA
VGs=O
iDrnA
-I
-2
-3
2 3 4 5=V
po
Vos
Fig 4.28 JFET drain characteristics.
4.7.3 EXPRESSION FOR PINCl-I OFF VOLTAGE, V
, p
Net charge in an alloy junction, must be the same,
semiconductor remains neutral.
e.N A Wp = e ND . wn
If NA » N
D
, from the above eq, wp « w
n
•
The relation between potential and charge density (p)
is given by Poisson's equation.
d
2
V e.N
D
dx
2
E
We can neglect w p and assume that the entire
harrier potential VB appears across the increased
donor ions. e.NI) = charge density P
d\' I!.N
n
,
:c- -----.x
e1x E
e.N
n
V=
E
At the boundary conditiuns, x = wn
_x
Fig 4.19 Charge density variation
~ l ~ ~ ~ ~ ~ ~ " ~
Fig 4.30 Structure
218
or
Wn = depletion region width
e.N
D
2
V = 2E . Wn
W =
n
1

e.N
D
Electronic Devices and Circuits
This is the expression for the penetration of depletion region into the channel. As the reverse
bias potential between Gate and drain increased, the channel width narrows or the depletion region
penetrates into the channel. Therefore the general expression for the spacecharge width
Wn (x) = w(n).
{
2 E }li
w(n) = -{Vo - V{x)}
eND
V = Voltage between drain and source V DS
w = Penetration of depletion region into the channel,
Where V = Vo - Vex); Vo is the contact potential at x and V (nl is the applied potential
across space charge region at x and is negative for an applied potential. This potential (between D
and S) is not uniform throughout the channel. The potential is higher near the drain and decreases
towards the source. The depletion region is more towards the drain and decrease towards the source.
.. V = Vo - V(x)
The actual potential is the difference of V 0 and V(x).
E = Dielectric constant of channel material
w(x) = a -b(x) = (Vo - V(x)>t ..... (1)
a - b(x) = Penetration w(x) of depletion region into channel at a point
x along the channel from one side of gate region only.
2a = Total width between the two gate diffusions.
2b(x) = Depletion region width between the two sides of the gate
regions.
If ID = 0, b(x) is independent of x and it will be uniform throughout the channel and will
be equal to 'b'. Vo will be much smaller than Vex).
If in equation (1), we substitute b(n) = b = 0, neglect V
o
, and solve for V we obtain the
pinch off voltage. Because at pinch off, the two depletion regions from both sides meet each
other, therefore the spacing between them b(x) = 0.
Therefore, w at pinch off, V 0 neglected, b = 0,

eND P
..
2,E •
or w
2
= clf.: .'v p
Q.
Transistor Characteristics
4.8 FET OPERATION
If a p-n junction is reverse biased, the majority carriers
will move away from the junction. That is holes on the
p-side will move away from the junction leaving negative
charge or negative ions on the p-side (because each atom
is deprived of a hole or an electron fills the hole. So it
becomes negative by charge). Similarly, electrons on the
n-side will move away from the junction leaving positive
ions near the junction. Thus, there exists space charge
on both sides of the junction in a reverse biased p-n
junction diode. So, the electric field intensity, the lines
219
of force originate from the positive charge region to the Fig 4.31 Space charge in p-n
negative charge region. This is the source of voltage drop ..
across the junction. As the reverse bias across the junction JunctIon.
increases, the space charge region also increases, or the region of immobile uncovered charges
increases (i.e., negative region on p-side and positive region on n-side increases as shown
in Fig. 4.31). The conductivity of this region is usually zero or very small. Now in a FET,
between gate and source, a reverse bias is applied. Therefore the channel width is controlled
by the reverse bias applied between gate and source. Space charge region exists near the
gate regi·on on both sides. The space between them is the channel. If the reverse bias is
increased, the channel width decreases. Therefore, for a fixed drain to source voltage the
drain current will be a function of the reverse biasing voltage across the junction. Drain
is at positive potential (for n-type FET). Therefore, electrons tend to move towards drain
from the source (1) Because, source is at negative potential, they tend to move towards
the drain. But because of the reverse bias applied to the gate, there is depletion region
or negative. charge region near the gate which restricts the number of electrons reaching
the drain. Therefore the drain current also depends upon the reverse bias voltage across
the gate junction. The term field effect is used to describe this device because the mechanism
of current control is the effect of the extensions, with increase reverse bias of the field
associated with region of uncovered charges
The characteristics of n-channel FET between I
D
, the drain current and V DS the drain
source voltage are as shown in Fig. 4.32, for different values of Vas.
0
- -
+0.5 V
J
OV
1
0
• rnA i ((
-2V
&
~ -3.5V
20V
----.. V OS' Volts
Fig 4.32 Drain charecteristics
To explain these characteristics, suppose Vas = 0. When Vos = 0, 10 ~ 0, because the
channel is entirely open. When a small Vos is applied (source is forward biased or negative
220 Electronic Devices and Circuits
voltage is applied to n-type source), the n-type bar acts as a simple semiconductor resistor and
so 10 increases linearly with Vos' With increasing current, the ohmic voltage drop between the
source and the channel region reverse biases the junction and the conducting portion of the channel
begins to constrict. Because, the source gets reverse biased or the negative potential at the
n-type source reduces. Because of the ohmic drop, along the length of the channel it self, the
constrictions is not uniform, but is wore pronounced at distances farther from the source. i.e.,
the channel width is narrow at the drain and wide at the source. Finally, the current will remain
constant at a particular value and the corresponding voltage at which the current begins to level
off is called as the pinch off voltage. But the channel cannot be closedown completely and
there by reducing the value of 10 to zero, because the required reverse bias will not be there.
Now if a gate voltage V G5 is applied in a direction to reverse bias the gate source
junction, pinch off will occur for smaller values of (V os), and the maximum drain current
will be smaller compared to when V GS = O. If V GS is made + O.SV, the gate source junction
is forward biased. But at this voltage, the gate current will be very small because for Si FET,
O.SV is just equal to or less than the cut in voltage. Therefore, the characteristic for
V GS = + O.SV, '0 value will be comparatively larger, (compared to V GS = OV or - O.SV). Pinch
off will occur early.
FET characteristics are similar to that of a pentode, in vacuum tubes.
The maximum voltage that can be applied between any two terminals of the FET is the
lowest voltage that will cause avalanche breakdown, across the gate junction. From the FET
characteristics it can be observed that, as the reverse bias voltage for the gate source junction
is increased, avalanche breakdown occurs early or for a lower value of Vos. This is because,
the reverse bias gate source voltage adds to the drain voltage because drain though n -type (for
n-channel FET), a positive voltage is applied to it. Therefore, the effective voltage across the
gate junction is increased.
For n-channel FET, the gate is p-type, source and drain are n-type. The source should
be forward-biased, so negative voltage is applied. Positive voltage is applied to the drain. Gate
source junction should be reverse biased, and gate is p-type. Therefore, voltage or negative
voltage is applied to the gate. Therefore, n-channel FET is exactly similar to a Vacuum Tube
(Triode). Drain is similar to anode (at positive potential), source to cathode and gate to grid,
(But the characteristics are similar to pentode).
For p-channel FET, gate is n-type, and positive voltage is applied, drain is at negative
potential with respect to source.
Consider n-channel FET. The source and drain are n-type and p-type gate is diffused
from both sides of the bar (See Fig. 4.33 below).
I
Gate
I
p
I
n - type channel
Source S Drain D
I
p
I
I
Fig 4.33 Structure of n-channel .I FET.
Transistor Characteristics 221
Suppose source and gate are at ground potential and small positive voltage is applied
to the drain. Source is IHype. So it is Forward biased. Because drain is at positive potential,
electrons from the source will move towards the drain. Negligible current flows between source
and gate or gate and drain, since these p-n junctions are reverse biased and so the current is
due to minority carriers only. Because gate is heavily doped, the current between Sand G or
G and D can be neglected. The current flowing from source to drain ID depends on the potential
between source and drain, V DS' resistance of the n-material in the channel between drain to
source. This resistance is a function of the doping of the n-material and the channel width;
length and thickness.
If V DS is increased, the reverse bias voltage for the gate drain junction is increased, since,
drain is n-type and positive voltage at drain is increased. This problem is similar to that of a
reverse biased p-n junction diode.
I ~ I -> I
_ 0 _ +
o _ +
r--::- ~ Pin:-+-
- I-0_--' __ ...L..-_--'-__ -.... +
I
+
I
v
Fig 4.34 Reverse biased G - D junctions
Consider a p-n junction which is reverse biased. The holes on the p-side remain near
the negative terminal and electrons on the n-side reach near the positive terminal as shown in
FigA.34. There are no mobile charges near the junction. So we call this as the depletion region
since it is depleted ofmobile carriers or charges. As the reverse voltage is increased, the depletion
region width 'I' increases.
This result is directly applicable for JFET. The depletion region extends more towards
drain because points close to the drain are at higher positive voltage compared to points close
to source. So the depletion region is not uniform V DS < V po. But extends more towards drain
than source. V PO is the pinch off voltage.
As V DS is increased, depletion region is increased (shaded portion). So channel width
decreases and channel resistance increases. Therefore, the rate at which ID increases reduces,
eventhough there is positive potential for the electrons at the drain and hence they tend to move
towards the drain and conventional current flows as shown by the arrow mark, in the FigA.35.
Source
Drain
Fig 4.35
222 Electronic Devices and Circuits
When V
DS
= Vpo
When Vos is further increased, the depletion region on each side of the channel join together as
shown in Fig.4.36. The corresponding V
DS
is called as VPD' the pinch off voltage, because
it pinches off the channel connection between drain and source.
Source Drain
Fig. 4.36
When Vos > Vpo
If V os is further increased, the depletion region thickens. So the resistivity of the channel increases.
Because drain is at more positive potential, more electrons tend to move towards the drain. Hence
10 should increase. But, because channel resistance increases, 10 decreases. Therefore, the net
result is 10 levels off, for any Vos above V po'
Vos
10= -
ros
In the linear region, ros is almost constant. So as Vos is increased, 10 increases. When
the two channels meet, as Vos increases r os also increases. So 10 remains constant.(See Fig.4.3 7)
Source Drain
Fig 4.37 Depletion region width variation in JFET.
4.9 JFET VOLT-AMPERE CHARACTERISTICS
Suppose the applied voltage Vos is small. The resulting small drain current 10 will not have
appreciable effect on the channel profile. Therefore, the channel cross-section A can be assumed
to be constant throughout, its length.
A = 2b.w,
Where 2b is the channel width corresponding to negligible drain current and w is channel
dimension perpendicular to the 'b' direction.
Io=Ae. No Il
n
E
A = 2b . w
Transistor Characteristics
E = VOS
L
VOS
2bwe . No Il
n
· L
Where L is the length of the channel. Eliminating 'b' which is unknown,
V GS = (
1
- r . V p
= 1 _ [VGs)7;
a Vp
Substituting, this value in equation (1 ),
ID = n V
DS
223
..... (1)
This is the expression for 10 in terms of V GS' V p and Vos because the value of b is not
directly known.
_ e.N
o
2
Vp - 28 . W
But, W = a - b (x)
b = 0, at pinch off,
. . w = a = The spacing between the two gate dopings.
.. [!Vp! = ;:-. a
2
[
VOS controls the width of the depletion region (a - b) because for a given V GS' as Vos
increases, the reverse potential between drain and gate increase. Therefore, depletion region
width increases.
EXPRESSION FOR V Gs
e.N
D
x
2
V=
2E
We can get the expression for V GS if we replace x by (a - b) and V by V GS'
e.ND(a - b)2
V GS = ---""-2'-E--'--
224 Elect."onic Dc, ices and Circuits
e.ND .a 2
IVpl =
2E
But
e.N
D
= IVpi
2E a
2
Substituting this in expression for V Gs'
Vp 2
V = -(a-b)
GS a2
Channel width is controlled by V GS' As V GS increases, channel width reduces.)
Therefore, channel width is controlled by V GS and depletion region width by V DS'
hs ~ (I-Hvp I
THE ON RESISTANCE r ds (ON)
When Vos is small, the FET behaves like an ohmic resistance whose value is determined by V GS'
The ratio
ID = 2bwe ND ~ n ' Vos
L
L
= r
2a we N D Iln ds(ON)'
4.10 TRANSFER CHARACTERISTICS OF FET
--+Vos
Fig 4.38 Drain characteristics
los = The saturation value of the drain current or the value at which ID remains
constant.
loss = The saturation value of the drain current when gate is shorted to source that
isVGs=O.
It is found that the transfer characteristics giving the relationship between los and V GS can
be approximated by parabola.
_ [1- VGS]2
los -loss Vp
Transfer characteristic follows the equation, ID vs V GS
[
VGS]2
ID = (loss + IGSS) 1- Vp - IGSS
With negligible I
GSS
' because IGS is the leakage gate current between G and S when Dis
shorted, to S will be of the orderofnano amperes, so it can be neglected.
Transistor Characteristics 225
_ (1- VGS]2
10 - loss Vp
This V p is V GS(oft) and not Vos because when V p = V GS(Oft),lo = O.
loss = The saturation value of drain current when gate is shorted to source or
VGs=O.
The transfer characteristics can be derived from the drain characteristics.
a
b
c
V
d
p
~
IOV
V
-4 -3 -2 -I
0
---+
Vos
Fig 4.39 Drain and Gate characteristics of JFET.
To construct transfer characteristics, a constant value of VOS is selected. Normally this
is chosen in the saturation region, because the characteristics are flat. The intersection of the
vertical line abed gives a particular value of 10 for different values of V GS' SO the transfer
characteristics between 10 and V GS can be drawn.
If the end points V p and loss are known, a transfer characteristics for the device can
be constructed.
Here V p is again called the pinch off voltage or V GS(oft) voltage. This is the voltage
between gate and source for which 10 becomes zero. As the reverse potential between G and
S increases, depletion region width increases. So channel width decreases. For some value of
V GS channel pinches off or 10 practically becomes zero (It cannot be exactly zero but few nano
- amps). So this gate source voltage at which 10 becomes zero is also called as pinch off voltage
or V GS(oft) voltage. This pinch off voltage is different from Vos voltage at which 10 levels off.
S i n c ~ in the latter case 10 is not zero, but channel width becomes zero and channel pinches
off. This channel width is made zero either by controll ing Vos or V GS' Hence there are two
types of pinch of voltages. The specified pinch off voltage V Jl for a given FET can be known
if it is due to Vos or V GS from the polarity of the voltage V . For n-channel FET, V GS is
negative (since gate is p-type, G-S junction is reverse biased) Vos is positive. If the specified
V p is -5V, (say) that it is due to V GS(off) because V p is negative. If V p is positive, it is
due to Vos at which loss levels off.
4.10.1 FET BIASING FOR ZERO DRIFT CURRENT
As Temperature (T) increases, Mobility (Jl) decreases at constant Electric Field (E), therefore
10 increases. As T increases depletion region width decreases. So conductivity 0' of the channel
increases, 10 also increases.
10 decreas.es by 0.7 % per degree centigrade.
226 Electronic Devices and Circuits
10 increase at a rate of 2.2 mVJOC change in V GS' Therefore change in V GS = 2.2 mVJOc.
. . For zero drift current,
0.007 11
0
1 = 0.0022 gm
or W=O.314 V
gm
or
IIVpl-IVGSI=0.63VI
(
VGS 12
Ios=loss 1- Vp )
..-- v
GS
g = g (1- VGS]
m mo V
p Fig 4.40 Biasing for zero drift current.
-
2
Ioss
~ o = V
p
4.14 Problem
Design the bias circuit for zero drain current drift, for a FET having the following particulars:
Vp =-3V; gmo = 1.8 mAN
loss = 1.75 rnA, if Rd = 5kn
(a) Find 10 for zero drift current.
(b) V GS
(c) Rs
(d) Voltage gain with Rs bypassed with a large capacitance,
Solution
For zero drift current,
(a)
(b)
(c)
(d)
IV pi-IV Gsl = 0.63
Vp =-3V
VGs=Vp-O.63
_ (1- VGS]2
lo-Ioss Vp :::::0.08 rnA
V GS =-3 -0.63 =-3.63 V
- V. 363 3.63
R = ~ = - ' - = =43Kn
s 10 10 0.08 rnA
V
GS
- Vp
gm = gmo V = 0.378 mAN
p
Ay = gm . Rd = 1.89
Transistor Characteristics 227
In
ohmic saturation breakdown
t
region region region
/VGs=O
'1'0
.............. .... ' ....
BV
DSS
VI'O --+ Vos
Fig 4.41 Drain characteristics.
The drain current 10 increases rapidly as Vos increases, towards V po' Above V po'
The current tends to level off at Ipo and then rises slowly. When V GS = Breakdown voltage
BV oss, breakdown (avalanche) occurs and the current rises rapidly. Current levels off because
as V increases, resistance of the channel also rises. Therefore, I remains constant. As V
increases, R increases, therefore % = constant. The region before V po is called ohmic region
because ohms law is obeyed (see Fig.4.41). Just as a p-n junction breakdown when reverse
voltage is very high FET also break down if V os is high due to high electric field.
Suppose Vos is fixed and V Gs is varied. As V GS is made negative, p-n junction is reverse
biased and depletion region between gate and source increases. This decreases channel width
and increases channel resistance. Hence 10 decreases. If gate voltage is made positive, depletion
region decreases and hence 10 increases. The p-n junction, between gate and source becomes
forward biased and current flows from gate to source. So n-type JFET is usually operated so that
V GS = 0 or negative.
Vpo
loss
BVoss
IO(oft)
IGSS
Pinch offvoltage when V GS = O.
o indicate V GS = O.
Saturation value of the drain current when gate is shorted to source
(V GS=O).
Saturation value of the drain current for V GS *- O. -I, -2 etc.
Breakdown voltage V
os
' when gate is shorted to source.
Drain current when the JFET is in OFF state.
Gate cut-off current when Drain is shorted to source.
Mutual conductance gm when V GS = O.
E.N
o
2
The expression for the pinch offvoltage IV pi = 2;-.a
where No = Donor atom concentration
E = Permitivity E - E
o r
a = Effective width of Silicon bar
228 Electronic Devices and Circuits
4.15 Problem
For a p - channel Silicon FET, with a = 2 x 10-4 cm and channel resistivity p = Ion -cm. Find the
pinch offvoltage.
Solution
ForSi,
or
or
4.10.2 CUT OFF
e.NA 2
Vp= ~ . a (forp-channeIFET)
E = 12 eo flp = 500 cm
2
/v - sec.
I
0' = p::: pflp. e : : : ' ~ A fl
p
' e
1
e.N =--
A Pflp
10x500
e.NA=2x 10-
4
V = 2xl0-
4
(2xlO-
4
L
p 2x12(36nx10
11
t
=3.78V
A FET is said to be cut off when the gate to source voltage IV Gsl is greater than the pinch off
voltage V p' IV Gsl > IV pl. Under these conditions, because for n - channel FET, gate is p - type and
source is n - type, there will be some current called gate reverse current or gate cut off current,
designed as I
GS
' this is when drain is shorted to source. There exists some drain current under
these conditions also called ID(off)'
10 (off) and IGSS will be in the range 0.1 amps to few nano amps at 25°C, and increase by a
factor of 1000 at a temperature of 150°C. IGSS is due to the minority carries ofthe reverse biased
gate. Source junction. IO(oft) is due to the carriers reaching drain from source because channel
resistivity will not become infinity. Because drain is at positive potential for n-channel FET, more
electrons will reach the drain.
4.10.3 ApPLICATIONS
1. JFETs make good digital and analog switches. offresistance is very high.
2. They are used in special purpose amplifiers such as very high input resistance amplifiers,
buffers voltage follower etc.
3. It can be used as a voltage controlled resistance because resistivity of the channel varies
with Vos.
4.11 FET SMALL SIGNAL MODEL
The equivalent circuit for FET can be drawn exactly in the same manRer as that for a vacuum
tube. The drain current 10 is a function of gate voltage V GS and drain voltage (Vos)
10 = f(V GS' V
os
) ..... (1)
Transistor Characteristics
Expanding (1) by Taylor's series,
aiD I
a:v-
GS VI"
aID I
+ ---av- . V OS
OS Vas
In the small signal notation the incremental values can be assumed to be a.c quantities.
= I
d
; V GS = V gs ; V os = V ds.
1
Id = gm . V gs + rd . V ds
where,
I Id I
or VDS = V VOS
gs
The mutual conductance or transconductance.
229
Sometimes it is designated as Yc or gc the second subscript's' indicating common source
and y or g for forwards transadmittancJ or respectively.
4.11.1 DRAIN RESISTANCE OR OUTPUT RESISTANCE r D
avOs I
ro= m VGS
o
aVos I Vds I
::: VGS = -1- VGS
d
,
, ,
, ,
" , ,
" , ,
I" ,
,

I', '
,
I ',
,
I,
I
I
• Saturation ----"0. I
.+- . ---..1
regIOn
Fig 4.42 Drain characteristics of JFET.
region
230 Electronic Devices and Circuits
Drain conductance = Yos = output conductance for common source.
gm ofFET is analog to gm of vacuum tube.
ro ofFET is analog to rp of vacuum tube.
TYPICAL V ALlJES
av I
IlforFET =
av
GS
10
gm =O.I-IOmAN
rd=O.I-IMO
p=rd,gm
C = I-IOpf
gs
Cds = 0.1 - 10 pf
4.11.2 EQlJIVALENT CIRCUIT FOR FET
= V os I = V
ds
I
10 V 10
GS gs
The for drain current Id in the case of a FET is
I
Id=gm V
gs
+ -,Vds
rd
This is similar to the expression Ip of a triode. So the equivalent circuit can be drawn as a
current source in parallel with a resistance. Between gate and source the capacitance is C
gs
and
C
gd
is the barrier capacitance between gate and drain. Cds represents the drain to source capacItance
ot the channel. (See Fig. 4.43).
Gate -----r---1
,... -.l C
gd
C
gs
Drain
Cds
Source
Fig 4.43 JFET equivalent circuit
This is high frequency equivalent circuit because we are considering capacitances. In low
frequency circuit, Xc will be very large so C is open circuit and neglected.
FET is a voltage controlled device; by controlling the channel width by voltage, we are
controlling the current. Ifwe neglect the capacitances the equivalent circuit ofFET can be drawn
as given in Fig. 4.44. This is similar of a transistor h-parameter equivalent circuit.
Fig 4.44 Simplified equivalent circuit
Trans istor Ch aracteristics 231
Arrow mark is downwards in the current source &n V gs because the current (conventional current)
is flowing from drain to source for n-channel FET (Because electrons are the majority carriers).
4.16 Problem
For a p-channel silicon FET, with a= 2 x 10-4 cm (effective channel width) and channel resistivity
P = 1 cm. Find the pinch off voltage.
Solution
e.N
A

V p = - for p-channel FET.
ForSi, E = 12 EO f..Lp= 500 cm
2
/v-sec.
1
0-= P =pf..Lp.e=NAf..Lp·e
1 1
or
e. NA = Pf..L
p
= tOx500
or e.N
A
=2 x l0-4
2 x 10-4 (2 x lO-4t
V = =3.78V
p 2XI2(361tXI011t
4. J 7 Problem
For an n-channel silicon FET, with a = 3 x 10-4 cm and No = 10
15
electrons/cm
3
• Find (a) The
1
pinch offvoltage. (b) The channel halfwidth for V GS ="2 V p and 10 = O.
Solution
(a)
(b)
e.N
D
2
V =--.a
p 2E
1.6 x 10-
19
XlO
l5
X(3XIO-
4
)2
2 x 12 x (36 n x lOll t
= 6.8V
b=? for
b= )11
= (3 x II>') (I-Gr)
= 0.87 x 10-4cm
232
4.18 Problem
Show that for a JFET,
Solution
g
= _2_ i1oss.Ios
m I Vp I"
_ (1- vGsJ
-g V
m p
Electronic Devices and Circuits
where gmo is the transistor conductance fdr V GS = O.
But
or
4.19 Problem
_ 21
Dss
gmo-- V-
p
= _ 21 DSS [1- V GS)
gnl V V
p p
(
1
- J = /:ssS
gm =- 2Ioss . los
Vp loss
=
I Vp I
Show that for small values of V GS' compared with V p, the drain current is 10::::: lDSS + gmo V GS .
Solution
Transistor Characteristics 233
I = I + g V, [ .. g - -
2
I [)S5]
D DSS mo (JS . IllO - VI'
4.12 FET TREE
The input Z of JFET is much higher compared to a junction transistor. But for MOSFETs, the
input Z is much higher compared to JFETs. So these are very widely used in ICs and are fast
replacing JFETs.
SOLID STATE DEVICES
FET
Depletion type
1.....----.-_....1
only
N-channel P-channel
Depletion type
~
N-channel P-channel
Fig 4.45 FET Tree.
MOSFET
(IGFEl)
Enhancement type
~
N-channel P-channel
There are two types ofMOSFETs, enhancement mode and depletion mode. These devices,
derive their name from Metal Oxide Semiconductor Field Effect Transistor (MOSFET) or MOS
transistor (See Fig. 4.46). These are also known as IGFET (Insulated Gate Field Effect Transistor).
The gate is a metal ang,.!-,> insulated from the semiconductor (source and drain are semiconductor
type) by a thin oxide layer.
In FETs the gate source junctions is a p-n junction. ( If the source is n-type, gate will be
p-type). But in MOSFETs, there is no p-njunction between gate and channel, but rather a capacitor
consisting of metal gate contact, a dielectric ofSi0
2
and the semiconductor channel. [two conductors
separated by dielectric] . .It is this construction which accounts for the very large input resistance
of 10 I
0
to 10
15
0 and is the major difference from the JFET.
234
The symbols for MOSFETs are,
G
~ - + - - - - o B
S
n-channel MOSFET, depjetion type
Fig 4.46
Electronic Devices and Circuits
G
S
p-channel MOSFET, depletion type
Fig 4.47
Some manufacturers internally connect the bulk to the source. But in some circuits, these
two are to be separated (See Fig. 4.46). The symbol, if they are connected together is, given in
Fig.4.47.
for p - channel, the arrow will point outwards.
Bulk is the substrate material taken.
Advantages of MOSFETs (over JFETs and other devices)
I. High package density::: 10
5
components per square cm.
2. High fabrication yield. p - channel Enhancement mode devices require 1 diffusion and 4
photo masking steps.
But for bipolar devices, 4 diffusions and 8 - 10 photo masking steps are required.
3. Input impedance is very high Zn::: 10
14
0.
4. Inherent memory storage: charge in gate capacitor can be used to hold enhancement
mode devices ON.
5. CMOS or NMOS reduces power dissipation, micropower operation. Hence, CMOS
ICs are popular.
6. Can be used as passive or active element.
Active: As a storage device or as an amplifier etc.
Passive: As a resistance or voltage variable resistance.
7. Self Isolation: Electrical isolation occurs between MOSFETs in ICs since all p-njunctions
are operated under zero or reverse bias.
Disadvantage
1. Slow speed switching.
2. Slower than bipolar devices.
3. Stray and gate capacitance limits speed.
4.12.1 ENHANCEMENT TYPE MOSFET
n-channel is induced, between source and drain. So it is called as n-channel MOSFET.
For the MOSFET shown in Fig. 4.48, source and drain are n-type. Gate is Al (metal) in
between oxide layer is there. The source and drain are separated by 1 mil. Suppose the p
substrate is grounded, and a positive voltage is applied at the gate. Because of this an electric field
will be directed perpendicularly through the oxide. This field will induce negative charges on the
Transistor Characteristics 235
semiconductor side. The negative charge of electrons which are the minority carriers in the
p - type substrate form an inversion layer. As the positive voltage on the gate increases, the
induced negative charge in the semiconductor increases. The region below the Si0
2
oxide layer
has n-type carriers. So the conductivity of the channel between source and drain increases and so
current flows from source to drain through the induced channel. Thus, the drain current is enhanced
by the positive gate voltage and such a device is called an enhancement type MOS.
Source Gate Drain
Induced Channel
p ( substute )
(n - channel MOSFET)
Fig 4.48 n-channel MOSFET
The volt-ampere drain characteristic of an n-channel enhancement mode MOSFET are
as shown in Fig. 4.49.
i
ohmic or
non saturation
l
constant current or
saturation region
o 10 20 30
-+ Vos
(a) Drain characteristics
i 5
loss
I
-2 -1 -+ V
GS
(b) Gate characteristics
Fig 4.49
The current 10 for V GS :s 0 is very small being of the order of few nano amperes. As
V GS is made positive, the current 10 increases, slowly first and then much more rapidly with
an increase in V GS (Fig. 4.49(b). Sometimes the manufacturers specify gate, source threshold
voltage V GST at which 10 reaches some defined small value:: 1 O ~ A . For a voltage < V GST'
(V GS threshold) 10 is very small. IO(Ot-l) of MOSFET is the maximum value of 10 which
remains constant for different values of V os'
In the ohmic region, the drain characteristic is given by
_ Il
C
o·
W
2
1
0
- 2L [2(VGS-VT)VOS-Vos]
236 Electronic Devices and Circuits
where, /..l = Majority carriers mobility.
Co = Gate capacitance per unit area.
L = Channel length.
w = Channel width perpendicular to C
The ohmic region in the 10 vs Vos characteristic of FET or MOSFET is also known as
triode region, because like in a triode 10 increases with Vos. The constant current region is
known as pentode region because the current remains constant with Vos'
Vos
+
s G
+ + + +
p - Substrate
(n...-channel MOSFET)
D
Electric field lines of force
Induced
n - channel
Fig 4.50 MOSFET structure
4.12.2 MOSFET OPERATION
Consider a MOSFET in which source and drain are of n-type. Suppose a negative potential is
applied between gate and source. Si0
2
is an insulator. It is sandwiched between two conducting
regions the gate (metal) and the S.C p-type substrate. Therefore, equivalent capacitor is formed,
with Si0
2
as the dielectric whenever a positive charge is applied to one plate of a capacitor a
corresponding negative charge is induced on the opposite plate by the action of the electric field
with in the dielectric. A positive potential is applied to the gate. So a negative charge is induced on
the opposite plate by the action of electric field within the dielectric, in the p - substrate. This
charge results from the minority carriers (electrons) which are attracted towards the area below
the gate, in the p - type substrate. The in the p - substrate are attracted towards the
lower region of Si0
2
layer because there is positive field acting from the gate through Si0
2
layer,
because of the applied positive potential to the gate V GS' As more number of electrons are attracted
towards this region, the hole density in the p - substrate (below the Si0
2
layer between n - type
source and drain only) decrease. This is true only in the relatively small region of the substrate
directly below the gate. An T} - type region now extends continuously from source to drain. The n-
channel below the gate is said to be induced because it is produced by the process of electric
inductor. If the positive gate potential is removed, the induced channel disappears.
If the gate voltage is further increased, greater number of negative charge carriers are
attracted towards the induced channel. So as the carriers density increases, the effective channel
resistance decreases, because there are large number of free carriers (electrons). So the resistance
seen by the Vos depends on the voltage applied to the gate. The higher the gate potential, the
lower the channel resistance, and the higher drain current 1
0
, This process is referred to as
Transistor Characteristics 237
enhancement because ID increases, and the resulting MOSFET is called enhancement type
MOSFET. The resistance looking into the gate is high since the oxide is an insulator.
The resistance will be of the order of 10
15
0 and the capacitance value will also be large.
p - Su bstrate
Fig 4.51 MOSFET biasing.
Depletion type MOSFET can also be constructed in the same way. These are also known
as DE MOSFETs. Here as the gate voltage increases, the channel is depleted of carriers. So
channel resistance increases.
Here the region below the gate is doped n - type. If negative voltage is applied to the gate,
negative charge on the gate induces equal positive charge i.e., holes. These holes will recombine
with the electrons of the n - channel between sources drain since channef resistivity increases.
The channel is depleted of carriers. Therefore ID decreases as V GS increases. (negative voltage)
Ifwe apply voltage to V GS' then this becomes enhancement type.
4.12.3 MOSFET CHARACTERISTICS
There are two types ofMOS FETs,
1. Enhancement type
2. Depletion type.
Depletion type MOSFET can also be used as enhancement type. But enhancement
type cannot be used as depletion type. So to distinguish these two, the name is given as depletion
type MOSFET which can also be used in enhancement mode.
As V GS is increased in positive values (0, + 1, +2 etc) ifiD increases, then it is enhancement
mode of operation because ID is enhanced or increased.
As V GS is increased in negative values (0, -I, -2, etc) if ID decreases, then it is depletion
mode of operation.
Consider n - channel MOSFET, depletion type. Fig. 4.52.
D
G

S
Fig 4.52 n - channel MOSFET.
238 Electronic Devices and Circuits
DRAIN CHARACTERISTIC : ID versus V DS
When Vas = OV, for a given V DS' significant current flows just like in a JFET. When the gate is
made negative (i.e., Vas = -1 V,) it is as if a negative voltage is applied for one plate of plate
capacitor. So a positive charge will be induced below the gate in between the n type source and
drain. Because it is semiconductor electrons and holes are induced in it below the gate. The
channel between the source and drain will be depleted of majority carriers electrons, because
these induced holes will recombine with the electrons. Hence, the free electron concentration in
the channel between source and drain decreases or channel resistivity increases. Therefore current
decreases as Vas is made more negative i.e., -1, -2, etc. This is the depletion mode of operation,
bt!cause the channel will be depleted of majority carriers as Vas is made negative (for n-channel MOSFEn.
In a FET, there is a p - n junction between gate and source. But in MOSFET, there is no
such p - njunction. Therefore, positive voltage positive Vas can also be applied between gate and
source. Now a negative charge is induced in the channel, thereby increase free electron
lornA
+2V
t
+IV
OV
-IV
Vas=-2 V
IOV 20V
Fig 4.53 Drain characteristics (N MOS).
concentration. So channel conductivity increases and hence ID increases. Thus, the current is
enhanced. So, the device can be used both in enhancement mode and depletion mode.
In the transfer characteristic as Vas increases, V D increases. Similady in the depletion
mode as Vas is increases in negative values, ID decreases
Depletion
+-
-I o
EnhancelOent
-+
Fig 4.54 Gate characteristics
Transistor Characteristics 239
A JFET is a depletion type, because as V GS is made positive (+ 1, +2 etc for n-channel FET)
ID increases. So it can be used in enhancement type. But there is no other type of JFET which is
used only in enhancement mode and no depletion mode, hence no distinction is made.
4.12.4 MOSFET GATE PROTECTION
The Si0
2
layer of gate is extremely thin, it will be easily damaged by excessive voltage. If the
gate is left open circuited, the electric field will be large enough (due to accumulation of charge)
to cause punch through in the Si0
2
layer or the dielectric. To prevent this damage, some MOS
devices are fabricated with zener diode between gate and substrate. When the potential at the
gate is large, the zener will conduct and the potential at the gate will be limited to the zener
breakdown voltage. When the potential at the gate is not large, the zener is open circuited and has
no effect on the device.
Ifthe body (bulk) ofMOS transistor (MOSFET) is p - type Silicon, and if two 'n' regions
separated by the channel length are diffused into the substrate to form source and drain n-channel
enhancement device (designated as NMOS) is obtained. In NMOS, the induced mobile channel
surface charges would be electrons.
Similarly, if we take n -type substrate and diffuse two p - regions separated by the channel
length to form source and drain, PMOSFET will be formed. Here in the induced mobile channel
surface charges would be holes.
4.12.5 COMPARISON OF p-CHANNEL AND n-CHANNEL MOSFETs
Initially there were some fabrication difficulties with n-channel MOSFETs. But in 1974, these
difficulties were overcame and mass productions ofn-channel MOSFETs began. Thus NMOSFETs
have replaced PMOSs and PMOSFETs have almost become obsolete.
The hole mobility in Silicon at normal fields is 500 cm
2
/v.sec. Electron mobility =
1300cm
2
/v-Sec. Therefore p-channel ON resistance will be twice that of n-channel MOSFET
ON resistance (ON resistance means the resistance of the device when 10 is maximum for a
1
given Vos) ON resistance depends upon' J.l' of carriers because (J = p = neJ.l
n
or peJ.l
p
'
If the ON resistance of a p-channel device were to be reduced or to make equal to that of
n - channel device, at the same values of 10 and V DS etc, then the p-channel device must have
more than twice the area of the n-channel device. Therefore n - channel devices will be smaller
or packing density of n-channel devices is more (R = ~ .R is decreased by increase in A).
The second advantage of the NMOS devices is fast switching. Tht! op·erating speed is
limited by the internal RC time constant of the device. The capacitance is proportional to the
junction cross sections.
The third advantage is NMOS devices are TTL compatible since the applied gate voltage
and drain supply are positive for an n - channel enhancement MOS.
(Because in n-channel MOS, source and drain are n-type. So drain is made positive.
For enhancement type the gate which is AI metal is made positive).
240 Electronic Devices and Circuits
4.12.6 ADVANTAGES OF NMOS OVER PMOS
I. NMOS devices are fast switching devices since electron mobility is less than holes.
2. NMOS devices are TTL compatible since V GS and V D to be applied for NMOS devices
are positive
3. Packing density of NMOS devices is more
4. The ON resistance is less because conductivity of NMOS devices is more since )..t of
electrons is greater than that of holes.
4.13 THE DEPLETION MOSFET
In enhancement type MOSFET, a channel is not diffused. It is of the same type (p-type or n-type)
as the bulk or substrate. But if a channel is diffused between source and drain with the same type
.of impurity as used for the source and drain diffusion depletion type MOSFET will be formed.
S G D
p - Substrate
Fig 4.55 Depletion MOSFET.
The conductivity of the channel in the case of depletion type is much less compared to
enhancement type. The characteristics of depletion type MOSFETs are exactly similar to that
of JFET.
When V DS is positive and V GS = 0, large drain current denoted as I
Dss
(Drain to source
current saturation value) flows. If V GS is made negative, positive charges are induced in the
channel through Si0
2
of the gate capacitor. But the current in MOSFET is due to majority carriers.
So the induced pesitive charges in the channel reduces the resultant current IDS' As V GS is made
more negative, more positive charges are induced in the channel. Therefore its conductivity further
decreases and hence ID decreases as V GS is made more negative (for n-channel depletion type).
,The current ID decreases because, the electrons from the source recombine with the
induced positive charges. So the number of electrons reaching the drain reduces and hence
ID decreases. So because of the recombination of the majority carriers, with the induced
charges in the channel, the majority carriers will be depleted. Hence, this type of
MOSFET is knows as depletion type MOSFET. JFET and depletion MOSFET have
identical characteristics.
A MOSFET of depletion type can also be used as enhancement type. In the case of
n-channel depletion type (source and drain are n-type), if we apply-positive voltage to the
gate-source junction, negative charges are induced in the channel. So, the majority carriers (electrons
in the source) are more and hence ID will be very large. Thus, depletion type MOSFET can also be
used as enhancement type by applying positive voltage to the gate (for n-channel type).
Transistor Characteristics
Problem 4.20
t
+4V
}
Enhancement
__ ---+-2------ Type
-2
-4
~ VDSv
Depletion
Type
Fig 4.56 Drain characteristics of DMOSFET.
~ Depletion 16
ID 12
t
-4 -2 2
~ Enhancement
Type
4
Fig 4.57 Gate characteristics of depletion type MOSFET.
241
A JFET is to be connected in a circuit with drain load resistor of 4.7 kn, and a supply voltage of
V DO = 30Y. V 0 is to be approximately 20V, and is to remain constant within ± 1 Y. Design a
suitable self bias circuit.
Solution
V 0 = y DO - Io . RL
I = Voo - Vo
o RL
30Y -20Y
4.7kn
= 2.1 rnA
for V 0 to be constant, it should be within ± 1 Y,
±V
,Mo= -R
. L
±IV
= 4.7kn ::: ± O.2mA
G
Fig 4.58 For Problem 4.20.
242
10 = (2.1 ± O.2mA)
Io(min) = (2.1 - 0.2) = 1.9mA
10 (max) = (2.1 + 0.2) = 2.3mA
Electronic Devices and Circuits
Indicate the points A and B on the maximum and minimum transfer characteristics of the
FET. Join these two points and extend.it till it cuts at point C.
The reciprocal of the slope ofthe line gives Rs
c
-+--VGS 1 --.v
GS
~ 1 ~ ~ - - - - - - - ~ V - - - - - - - - - - ~ ~ ~ 1
Fig 4.59 For Problem 4.20.
~ V lOY
Rs = ~ I = 2.5mA = 4kO
The bias line intersects the horizontal axis at V G = 7V
An external bias of 7V is required.
R2
VG= .Voo
R
J
+R2
R2 7
R
J
- 23
R2 and R\ should be large to avoid overloading input signals. IfR2 = 700 kO,
R\ =2.3 MO
4.21 Problem
(a) For the common source amplifier, calculate the value of the voltage gain, given
rd = 100 kO, gm = 300 J.I. , RL = 1O.kO, Ro = 9.09kO
-3000xl0-
6
x 100 x 10
3
xlOxl0
3
Solution
(100 x 10
3
)+ (lOX 10
3
)
=-27.3
Transistor Characteristics 243
(b) If Cos = 3pf, determine the output impedance at a signal frequency of 1 MHz.
Solution
X = 1
c 21tf Cos
f= 1 MHz,
1
X = = S3kn
c 21txlxI06x3x10-12
IZol = 9.09knx S3kn = 8.96kn
+ (S3kn)2
4.14 CMOS STRUCTURE (COMPLEMENTARY MOS)
This is evolved because of circuits using both PMOS and NMOS devices. It is the most sophisticated
technology. CMOS devices incorporate p-channel MOSFET and n-channel MOSFET.
The advantages that we get with this complementary use of transistor are:
1. Low power consumption.
2. High speed.
But the disadvantages are
I. Fabrication is more difficult. More number of oxidation and diffusion steps are involved.
2. Fabrication density is less. Less number of devices per unit area, because the
device chip area is more.
4.14.1 MNOS STRUCTURE
In this process, we will have Silicon semiconductor. l;)i0
2
over it and then layer of Silicon Nitride
(Si
3
N
4
) and a metal layer. So we get the Metal, Nitride, (Si
3
N
4
) Oxide (Si0
2
) and Semiconductor
structure. Hence, the name MNOS. So the dielectrics between metal (AI) and semi-conductor
(Si) is a sandwich ofSi0
2
and Si
3
N
4
, whereas in ordinary MOSFETs we have only Si0
2
.
AI AI Al
N - Si
Fig 4.60 MNOS FET.
244 Electronic Devices and Circuits
The advantages are :
1. Low threshold voltage V T. (V T is the voltage of the drain, (V 0) beyond which only the
increase in drain current 10 will be significant)
2. Capacitance per unit area of the device is more compared to MOS structure. Because
the dielectric constant will have a different value since because C is more, charge storage
is more. It is used as a memory device.
4.14.2 SOS - MOS STRlJCTURE
This is Silicon On Saphire MOS structure. The substrate used (Fig.4.61) is the silicon crystal
grown on Saphire subtrate. For such a device, the parasitic capacitance will be very low.
A/ Al Al
+
n p
n
+
p
Insulating saphire substrate
Fig 4.61 SOS - MOS Structure.
4.14.3 VOLTAGE BREAKDOWN IN JFETs
n
There is a p-n junction between gate and source and gate and drain. Just as in the case of a
p-n junction diode if the voltage applied to the p-n junction of G and S, or G and D junction or
D and S junctions increase, avalanche breakdown will occur.
Io
Fig 4.62 JFET circuit.
BV
DGO
:
This is the value of V DG that will cause junction breakdown, when the source is left open i.e., Is = 0
BV
GSS
:
This is the value of V GS that will cause breakdown when drain is shorted to source. When
drain is shorted to source, there will not be much change in the voltage which causes breakdown.
So, BV GSS = - BV DGO
Transistor Characteristics 245
For a p-channel FET, Gate is l1-type, Sand Dare p - type. A positive voltage is applied to
G and negative voltage to drain. Therefore with respect to D. Y OG is negative voltage. Y GS is
positive voltage because G is at positive potential with respect to source. Since BY GSS = - BY OGO
gate junction breakdown can also result by the application of large Y Oli'
Yos = Y OG + Y GS
More Y
os
that can be applied to a FET = Max. Y OG + Max. Y GS
BY
os
X = Breakdown voltage Yos for a given value of Y GS'
X indicates a specific value ofY GS
BY osx = BY DCiO + Y GSX
4.22 Problem
Determine the values ofY GS' '0' and Yos for the circuit shown in using the following data.
loss = 5mA; Y p = - 5 Y; Rs = 5kO; RL = 2kQ; Y 00 = lOY
Solution
The gate is at DC ground potential, since no DC voltage is applied between gate and Source.
Y GS + Is . Rs = 0
But Is::: 1
0
, IG is negligible
Y GS + 10 Rs = 0
or Y GS = - 10 . Rs
= - 5000 1
0
, Vo
Y GS is unknown and 10 is unknown.
But
?
10 = 10SS(I- ~ c ~ ~ r
=5 x 10-
3
(1- ~ G 5 S r
Y
GS
10 = - 5000
V GS = (-5000)(5 x 10-
3
)(1 + 0.2Y Gsi,
Y GS
2
+ 11 Y GS + 25 = 0
Y GS = - 3.2V and V GS = -7.8Y.
c
Fig 4.63 For Problem 4.22
The pinch off voltage V p for which 10 becomes zero is given as 5V therefore V GS cannot be
7.8 Y. Therefore Y GS = - 3.2Y.
10 = (5 x 10-
3
)( I - 0.64)2 = 0.65 mA
IlJ' RL = 2000 10 = 1.3Y
1
0
, Rs::: 5000 10 = 3.25Y
Vos = 10 - 1.3 - 3.25 = 5.45Y.
The FL'; must be conducting.
If Y GS = - 7.8Y. the FET in cut off. Therefore Y p = -5Y. Therefore Y GS is chosen
as - 3.2Y.
246 Electronic Devices and Circuits
4.15 SILICON CONTROLLED RECTIFIER
Silicon controlled rectifier (SCR) is a 4 layer p-n-p-n device. In one type of construction, n-type
material is diffused into a silicon pnp pellet to form alloy junction. From the diffused n region, the
cathode connection is taken. From the p region on the other side anode is formed. Gate is taken
from the p - region into which n - type impurity is diffused. So in the symbol for SCR, the gate is
shown close to the cathode, projecting into it. This is known as planar construction.
Gate
p
K
fG
n
p
A
Anode
Fig 4.64 SCR structure. Fig 4.65 SCR symbol.
4.15.1 ANNULAR TYPE OF CONSTRUCTION OF SCR
In this type of construction gate is central to the device, with the cathode surrounding it. This is
also called as shorted emitter construction. The advantages of this type construction are
(i) Fast turn on time
(ii) Improved high temperature capabilities.
4.15.2 STATIC CHARACTERISTICS
The gate current Ia determines the necessary forward voltage to be applied between anode and
cathode (Anode positive with reference to cathode) to cause conduction or to turn SCR 'ON'. So
that current flows through the device.
As the gate current increases, the voltage to be applied between anode and cathode to turn
the device ON decreases. If the gate is open, Ia = 0 , the SCR is in the OFF state. This is also
known as forward blocking region. The anode current that flows is only due to leakage. But if
the voltage between anode and cathode is increased beyond a certain voltage, V BO called as the
forward breakover voltage, the SCR will turn ON and current will be limited by only the applied
voltage V BO called as the forward breakover voltage, the SCR will turn ON and current will be
limited by only the applied voltage.
V BOO: THE FORWARD BREAK OVER VOLTAGE
This is the voltage to be applied between anode and cathode to turn the SCR ON, when Ia = O.
B.O : Breakover 0 : Zero gate current.
As the gate current is increased, the forward break over voltage decreases. If the value of
Ia is very large ;;:: 50 rnA, the SCR will turn ON immediately when same voltage is applied
oetween anode and cathode.
Transistor Characteristics 247
Usually, some voltage is applied between the anode and cathode of the SCR, and it is turned
ON by a pulse of current in the gate. S
IA
I ~ ___ High conduction
region ON state
•
Holding <:urrent IH
-
~ ~ .. ~ ~ ~ ~
Leakage current
I = 0
G
Fig 4.66 SCR Characteristics
HOLDING CliRRENT IH
Once SCR is ON, a minimum amount of current is required to flow to keep the device ON. Ifthe
current is lowered below IH' by increasing external circuit resistance, the SCR will switch off.
Once the SCR is turned ON, the gate looses control over the device. The gate can not be
used to switch off the device. If V A _ K is reduced to zero ( as automatically happens in rectifying
application) or if the current is reauced below I If the device will turn off.
IH : Holding Current
Minimum Anode Current required to keep SCR in the conducting state. IfIA goes below I
H
, SCR
is turned off ( To hold in conducting state ).
LATCHING CliRRENT IL
Minimum value of I A to be reached to keep SCR in the ON state even after the removal of gate
trigger signal. (To latch on to the conducting state.)
4.15.3 ANALYSIS OF THE OPERATION OF SCR
The SCR can be regarded as two transistor connected together.
The SCR can be regarded as two transistor p-n-p and n-p-n. The Base of p-n-p transistor is
connected to the collector of the n-p-n transistor. The base ofn-p-n transistor is connected to the
collector of p-n-p transistor. The emitter of n-p-n transistor is the cathode. The emitter of p-n-p
transistor is the anode.
Suppose a positive voltage is applied to the anode (p type) since there is no base injection for
both the transistors, the transistors are off. The only currl';)nt is the leakage current. leOI and le02
of the two transistors. Therefore when SCR is not tUrI'!cd ON, the current that results is the
leakage current (leOl + le02)'
248 Electronic Devices and Circuits
p n
n
B2 P
P G E2
'----.-----'
n
K
Q
1
le2l
r3
1
leo1
leo2
r3
2
IG
G
Fig 4.67 Two transistor analogy of SCR.
In a transistor,
CONSIDER TRANSISTOR Q
1
The base current for the transistor Q I' IB I is the collector current IC2 of Q2'
Ic = r3 (IB + lco ) + Ico
IBI = IC2
ICI = r3
I
(IC2 + ICOI ) + ICOI
CONSIDER TRANSISTOR Q
2
The base current of transistor Q
2
is the collector current Ic I of Q I'
IC2 = [32 ( ICI + IC02 ) + IC02 .
Anode current is IE2 = IB2 + IC2
.......... ( 1)
.......... (2)
.......... (3)
.......... (4)
.. ........ (5)
Transistor Characteristics
But IB2 = IC1
. IA = IC1 + IC2
I A can be written in the forms,
. - - - - - - - - - - - - - - - - - ~
(1 + PI Xl + P2 XI
col
+ I
C02
)
J -PI' P2
249
The two transistor will have wide base regions. So 13 will be small. The value of 13 depends
upon the value oflE' When IE is very small, 13 ::: O.
Therefore when 13
1
= 13
2
= O.
(lcol +IC02Xl+0Xl+0)
IA = 1-0.0
::: ICOJ + IC02
When a negative voltage is applied from gate to cathode, to inject holes into the base ofO]'
emitter base junction of0
1
is forward biased. So leI increases. But this is the base current for 02'
So transistor 02 is tuned uN. This increases collector and emitter currents. The IC2 of 02 now
becomes trigger current and so increase I
C1
' So the action is one of internal regeneration feed
back, until both the transistor are driven into saturation. Once the SCR is turned ON, removal of
gate current will not stop conduction because already the transistor 02 is turned ON and it provides
the base injection for ° ,.
SCR can be represented as a pnp and npn transistors connected together, as shown in
Fig. 4.68.
P E)
IB)=IC2
C
2
n
B)
le2l
13) leo)
B2
p
G
C)
leo2
13
2
E2
n
IB2 IG
G
K
Fig 4.68 Two transistor analogy of SCR.
250
IBI = IC2
ICI = IB2
Electronic Devices and Circuits
Anode current = IA = lEI
General expression for Ic in term of f3, IB and Ico is
Ic = f3 IB + (f3 + 1) Ico
Since for transistor QI'
ICI = f3
1
IBI + ( f3
1
+ I) ICOI .......... (6)
Since for transistor Q l'
IC2 = f3
2
IB2 + (f3
2
+ 1) IC02
But IB2 = ICI
. . IC2 = f3
2
Ic I + (f3
2
+ 1) ICo2 .......... (7)
IA = lEI = ICI + IBI But ICI is given by eq. (6)
= f3
1
. IBI + (f3
1
+ 1) ICOI + IBI
.. IEI=IA=(f31+1)(lBl+IcOl) .......... (8)
Since we must get an expression for IBI"
Substitute eq. (6) in (7).
.. IC2 = f3
2
[f3
1
I
BI
·+ (f3
1
+ 1) I
COl
] + (f3
2
+ 1) IC02
But IC2 = IBI
.. IBI = f3
2
f3
1
IBI + f3
2
(I + f3
1
) ICOI + (f3
2
+ 1) IC02
or (1- f3
1
f3
2
) IBI = f3
2
(1 + f3
1
) Icol + (1 + f3
2
) Ic02
132 (1 + PI )Icol + (1 + 132 )1 CO2
or IBI = .......... (9)
1-131132
Substitute the value ofIB I from eq. (9) in (8).
.. I =1 =(1+A )[132(l+13I)I
col
+(l+132)lc02 +1 1
EI A 1-'1 1-13
1
13
2
COl
Transistor Characteristics
4.15.4 ANODE TO CATHODE VOLTAGE - ClIRRE,\T CHARACTERISTIC
o - 7: Reverse blocking
characteristic
o - I, 0 - 6 : The device
o - 2: Forward blocking state
I - 2: Forward avalanche region
2: Forward breakover point
2 - 4: Negative resistance region
251
is in blocking state
3 - 5: Forward Conducting characteristic
3 : Holding current
2
...
. -.
. . -

__
1
6
V 80: Breakover voltage when IG = 0
7
Fig 4.69 SCR characteristics.
4.16 UNIJUNCTION TRANSISTOR (UJT)
This device has only one p-n junction and three leads like transistor. Hence it is called as unijunction
transistor (UJT).
The construction ofthis device is as shown in Fig. 4.70.
P-N junction

Aluminium rod ./'
Base n-typesilicon bar
B1
Ohmic
contacts
(a) UJT structure
(b) Symbol
Fig 4.70 (a) UJT structure (b) Symbol
252 Electronic Devices and Circuits
A bar of high resistivity n-type silicon of typical dimensions 8 x 10 x 35 mils called base
B, is taken and two ohmic contacts are attached to it at the two ends to form the base leads B2 and
B]. A 3 mil aluminium wire, called emitter E is alloyed to the base close to B2 to form a p-n
rectifyingjunction. This device was originally described in the literature as the double-base diode.
But now it is called as UJT (.: it has only one p-n junction and two base leads).
4.16.1 UJT CONSTRUCTION AND OPERATION
In UJT, the alloy is formed by diffusing Alluminium into n type silicon close to the B2lead as shown
in Fig. 4.71. The doping and construction is such that, when E - B],junction is forward biased,
holes are injected into the n-silicon bar from the p-region towards the B] lead. Holes will not travel
upwards because B2 is at positive potential. The doping concentration of p-region is large. So there
is sudden increase in the number of holes in the region close to B]. To recombine with these holes,
the free electrons from B2 region will mo"e towards the B] lead. Thus there is large increase in the
number of holes and electrons in the region corresponding to R
B
].
+v
E
n-Silicon bar
Fig 4.71 UJT structure.
:. (J == ne Il
n
+ p e II
p
' p will be almost comparable with n. Because of increase in the
values of nand p, (J .increases, R
B
, decreases. When the E-B, junction is forward biased, even
after recombination of holes and electrons, still there will be large number of electrons and holes in
the region close to B]. So conductivity (J increases.
RB and RB have the same temperature coefficient. Therefore material is the same
(n type sfIicon bar). With temperature, because of the increase in the number of free carriers,
RB decreases. The net :effect is n decreases slightly with the temperature. The change is :::: 4%
for 100
0
C rise in temperature. V p = (11 V BB + V v), V v decreases with increase in temperature.
Because threshold voltage decreases with increase in temperature. Therefore V p decreases with
temperature. Now R2 has positive temperature coefficient. If R2 is chosen such that it increases
by the same amount as to decrease in R
B
] and 11 and V v' V p will remain constant. Thus, R2 will
provide temperature compensation.
UJT basically consists of a bar of n-type silicon with one p-type emitter providing p-n
junction. The emitter will be close to the base two (B
2
) than base one (B]) (see Fig. 4.73). So
UJT consists of emitter and two bases with the emitter close to the second base. The emitter is p-
type and the two bases are n-type. The symbol for the voltage between emitter and base 1 (B] is
n-type). Polarity is as shown in Fig. 4.72. Suppose V
BB
is the voltage from B2 to B]. Now if
V BB == 0, and positive voltage V E is applied. The resulting current IE gives the emitter to base B]
diode characteristic. With V BB == 10V or 20V, there is leakage current lEO from B2 to emitter.
Transistor Characteristics 253
(.,' B2 is n-type-and E is p-type when B2 is at a higher potential, the minority carries from B2 and
E will flow which results in leakage current). It takes::::: 7 volts from E to B" to reduce this current
to zero and then cause current to flow in the opposite direction, reaching peak point V p' After this
point IE increases suddenly and V E drops. This is the unstable resistance region. This lasts until
the valley voltage is reached and the device saturates.
+
(a) Symbol (b) V-I components
Fig 4.72
The equivalent circuit for UJT is as shown in Fig. 4.73. The resistance of the silicon bar is
represented by two series resistors. RB is the resistance of base two portion, RB is variable
resistance. Its value depends upon the biJs voltage Yo' The p-n junction formed due Ito emitter is
shown as a diode.
RS2
i
+
i
V
BB
RBI
V
RSI
~ ~
V
RB
, = llV
BB
= R R x V
BB
, where,
B' + B2
RB
II = R + ~ = Intrinsic stand off ratio.
B, B2
Its value lies between 0.5 and 0.82. Since no voltage is applied at the emitter, the diode is
reverse biased, because the cathode voltage is II V BB and anode current is O. Now as V E is
increased, from zero and when V E is> II V BB' the diode will be forward biased. So the resistivity
between E and 8, decreases since holes are injected from emitter into 8,. Therefore R
B
, decreases or
V E decreases and IE increases. So negative resistance region results. Peak voltage V p = II V BB + V y'
Vy = O.SY.
254 Electronic Devices and Circuits
4.16.2 V - I CHARACTERISTIC OF UJT
Suppose some voltage V BB is applied between B2 and B
I
. If V BB is small, IE = O. Then the silicon
bar can be considered as an ohmic resistance RBB between the leads B2 and B
I
. RBB will be in the
range 5 to 10 kn. RBB = (RBI + R
B2
). When IE = 0, the voltage across RBI = 11VBB" Where,
11 = RBI .11 is called is intrinsic stand off ratio and its value lies between 0.5 and 0.75.
RBI + RB2
If V E is < 11 V BB the p-njunction diode or the E-B junction ofUJT is reverse biased and the current
IE is negative. It is the reverse saturation current lEO which is of the order of 10 R
B
.
If VEE is increased beyond 11 V BB' input diode becomes forward biased and becomes positive.
So holes are injected from the emitter into the base region B
I
. As VEE increases,
IE increases. So the holes injected into B1 region also increases (cr = pe II
p
).
:. As the hole concentration in the base region increases, its resistivity decreases. So the
voltage across R
B
, 11 V BB decreases. This is the voltage V E between E and B
I
. So as IE increases,
V
E
decreases and the device exhibits negative resistance region. (Fig. 4. 74(b))
When IE is very large the current IB2 flowing from V BB into B2 load can be neglected. For
current greater than the valley current, the resistance becomes positive.
V p = 11 V BB + V V
V p = Peak Voltage
V v = Valley Voltage
For current greater than the valley current the resistance becomes positive. (Fig. 4.74.)
B
I
(a)
. I
negative ~
resistance region :
I
: ~ Saturation
region
- V;------
I valley point
(b)
Fig 4.74 (a) UJT circuit (b) V-I Characterisitic of UJT
4.16.3 ApPLICATIONS OF UJT
1. In Relaxation oscillator circuits to produce sawtooth waveforms.
2. In triggering SCR circuits.
3. In thyristor circuits.
Transistor Characteristics
255
4.17 LED'S
These are Light Emitting Diodes. On the application of voltage, when electron transition occurs
from excited state to lower energy level, the difference in energy is released in the form of light.
Thus I ight emission takes place. Th is property is exhibited by direct band gap semi conductor materials.
The symbol for LED and its V-I characteristics are shown in FigA.75. LEDs are always
used in Forward Bias. Because when reverse biased energy transition and hence light emission
will not take place.
Direct Band Gap Materials: Conduction band minimum energy and valence band maximum
occur at the same value of K (wave vector) in the E - K diagram of semiconductors.
Indirect Band Gap Conduction band minimum and valence band maximum energy do not
occur at the same value of K (wave vector) in the E - K diagram.
Photons have high energy and low momentum.'
hC
/..= E; h ·f= E
I/.. = ~ AOI
18 the complete-electromagnetic spectrum visible spectrum extends in the range ::: 4000 -
7200 A . Semiconducting materials suitable for light emitters should have energy gap EG in the
range l. 7 5 - 3.15 e V.
Direct Band gap semiconductors usedfor LEDs: (GalliumArsenide)GaAs,(Gallium
Antimony) Ga Sb, (Arsenic) As, (Antimony) Sb, (Phosphorous) P.
Impurities added are Group II materials: (Zinc) Zn, (Magnesium) Mg, (Cadmium) Cd,
Group VI donars: Tellicum Te, Sulphur S.
Impurity concentration: 10
17
- 10
18
fcm
3
. for donor atoms.
Impurity concentration for acceptor atoms: 10
17
- 10 19 fcm
3
.
LEDs are based on Injection IIIuminisence. Due to injection of carriers light is emitted.
Gallium phosphide - Zinc oxide LED Red colour
Gallium phosphide - N LED Green colour
Silicon Carbide - SiC LED Yellow colour
Gallium phosphide, P, N LED Amber colour
4.18 PHOTO DIODES
Instead of photoconductive cells, if a reverse biased p-njunctions diode is used the current sensitivity
to radiation can be increased enormously. The mechanism of current control through radiation is
similar to that of a photo conductive cell. Photons create electron hole pairs on both sides of the
junction. When no light is applied the current is the reverse saturation current due to minority
carriers. When light falls on the device photo induced electrons and holes cross the junction and
thus increasing the current through the device. For Photo diodes also dark current will be specified.
This is the current through the device with no light falling on the device. It is typically 30 - 70 n A
at V
R
= 30 V.
The active diameter of these devices is only 0.1". They are mounted in standard To - 5
package with a window.
256 Electronic Devices and Circuits
They can operate at frequencies of the order of 1 MHz. These are used in optical
communication and in encoders.
A A
~
~ or
t ~
Light
I ~ J . L A
intensity
K K
Fig 4. 76 Symbols for photo diode.
4.19 PHOTO TRANSISTORS
Fig 4.77 Photo diode reverse characteristics.
In a photodiode, the current sensitivity is much larger compared to a photoconductive cell. But this
can be further increased by taking advantage of the inherent current multiplication found in a
transistor. Photo transistors have a lens, to focus the radiation or light on to the common base
junction of the transistor. Photo induced current of the junction serves as the base current of the
transistor (I,,).
or
Fig. 4.78 Symbols for Photo transistors
Therefore collector current Ie = ( 1 + h
fe
) 1,,-
The base lead may be left floating or used to bias the transistor into some area of operation.
Symbol for photo transistor:
Ifthe base lead in left open, the device is called photo duo diode.
Symbol is
It has two diodes pn and np. So it is called as photo duo diode.
Fig. 4.79 Symbol for Photo duo diode.
Transistor Characteristics 257
SUMMARY
• The relation between Emitter Efficiency y, Transportation Factor 13* and Current
Gain a is,
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
* .
a=13 xy
Transistor is an acronym for the words Transfer Resistor. As the input side in
forward biased and output side is reverse biased, there is transfer of resistance
from a lower value on input side to a higher value on the output side.
Transitor can be used as an amplifiers, when operated in the Active Region. It is also
used as a Switch, when operated in the cut-off and saturation regions.
The three configurations of Transistor are Common Emitter, Common Base and
Common Collector.
The proper name for this device being referred as transistor is Bipolar Junction
Transistor ( BJT ).
The three regions of the output characteristics of a transistor are
1. Active Region
2. Saturation Region
3. Cut-off Region
JFET is UNIPOLAR Device (Unipolar only one type of carriers either holes or
electrons)
JFET device has Higher input resistance compared to BJT and Lower input resistance
compared to MOSFET.
The disadvantage of JFET amplifier circuits is Smaller Gain - Bandwidth product
compared to BJT amplifier circuits.
avosl
r
ds
(ON) = aI
D V", = K
aVDSI
J.l = Amplification factor = av
GS In= K
Relation between J.l, rd and gm for a JFET is J.l = rd x gm
Width of the depletion region W n for n-channel JFET, interms of pinch off voltage V p
is W =
eND
Expression for 1m; in tenns of Inw V GS and V, is In Inss {I - r
I
Dss
= Satuaration value of Drain current when gate is shorted to source.
258
•
•
•
•
•
•
•
•
•
•
Electronic Devices and Circuits
For zero drift current in the case of JFET, 0.00711DI = 0.0022 gm
For zero drift current, in the case of JF ETs, IV pi-IV osl = 0.63V
Expression for gm for JFET interms of gmo is, gm = (1- ~ : s )
The two types ofn-channel or p-channel MOSFETs are
I. Depletion type
2. Enchancement type
In JFET terminology, B V DGO parameter means the value of breakdown voltage V DO
at breakdown when the source is left open i.e., Is = O.
The parameters B Voss stands for breakdown voltage Vos when drain is shorted to
source.
At high frequencies, the expressions for input capacitance C
m
of JFET shunting Ro
is, C
m
= C
gs
+ [I + gm (R
L
II r
d
)] C
gd
The purpose of swamping resistor rs connected in series with source resistance ~ in
common source (C.S) JFET amplifier is to reduce distortion.
Simplified expression for voltage gian Ay in the case of common Drain JFET amplifier
Rs
is A = ----"--
y 1
Rs+-
gm
The distortion caused due to the non linear trarfsfer curve of JFET device, in amplifier
circuits is known as Square Law Distorion.
OBJECTIVE TYPE QUESTIONS
I. For identical construction, .............................. types of bipolar junction transistors
(BJTs) have faster switching times. .
2. The arrow mark in the symbols of BJTs indicates ............................. .
3. For NPN Bipolar Junction Transistor Emitter Efficiency y = ............................ ..
4. For PNP Transistor (BJT), the transportation factor 13* = ............................. .
5. Relation between a, 13* and y for a transistor ( BJT ). a = ............................. .
6. The expression for a in terms of 13 for a BJT is ............................. .
7. The expression for 13 in terms of a for a BJT is ............................ ..
8. Expression for Ie in terms of 13, IB and leBO is ............................ ..
9. Expression for leEO in terms of leBO and a is ............................ ..
10. J3' of a BJT is defined as ............................. .
II. J3' of a transistor ( BJT ) is ............................ ..
12. Relation between J3' and 13 is ............................. .
13. J3' is synonymous to .............................. and 13 is same as ............................. .
14. 13 is .............................. ( for DC Currents ).
15. J3' is .............................. ( for AC Currents ).
Transistor Characteristics 259
16. Base width modulation is ............................. .
17. IGFET is the other name for .............................. device.
18. In JFET recombination noise is less because it is .............................. device.
19. The disadvantage of JFET amplifier circuit is .............................. .
20. The D, G, S terminals of JFET are similar to .............................. terminals of BJT
respectively.
21. The voltage V DS at which ID tends to level off, in JFET is called .............................. .
22. The voltage V GS at which ID becomes zero in the transfer characteristic of JFET is
called .............................. .
23. The range of r
DS
(ON) for JFET is .............................. .
24. For low electric fields E of the order of 10
3
V fcm, /J. ex .............................. .
x
25. Expression for V GS in terms of V P' .............................. .
26. Expression for IDX in terms of IDSS in IDS ............................. .
27. IDSS in defined as .............................. .
28. JFET can be used as .............................. resistor.
29. Relation between /J., rd and gm for JFET is .............................. .
30. The square law device is ............................. .
31. The MOSFET that can be used in both enhancement mode and depletion mode is
ESSAY TYPE QUESTIONS
1. With the help of a neat graph qualitatively explain the Potential distribution through <t
transistor (BJT).
2. Explain about the different current components in a transistor.
3. Derive the relation between 13 and 13' .
4. Derive the relation between ex, 13* and y
5. Differentiate between the terms hFE and h
fe
. Derive the relationship between them.
6. Qualitatively explain the input and output.
7. Explain the V-I characteristics in Common Emitter Configuration.
8. Describe the V-I characteristics of a transistor in Common Collector Configuration
and Explain.
9. Draw the Eber-Moll Model of a transistor for NPN transistor and explain the same.
10. Compare the input and output characteristics of BJT in the three configurations,
critically.
11. Compare BJT, JFET and MOSFET devices in all respects.
12. Derive the expression for the width of Depletion region' W' in the case of p-channel
JFET.
13. Obtain the expression for the pinch offvoItage Vp in the case ofn-channel JFET
14. Deduce the condition for JFET biasing for zero drift current.
"rl
260
15.
16.
17.
18.
19.
20.
Electronic Devices and Circuits
Draw the structure and explain the static Drain and Gate characteristics of n-channel
JFET. Repeat the same for p-channel JFET.
Draw the structure of p-channel MOSFET and Qualitatively explain the static Drain
and Gate characteristics of the device.
What are the applications of JFET and MOSFET devices?
Give the constructional features ofUJT.
Qualitatively explain the static V-I characteristics ofUJT.
What is the significance of negative resistance region? Explain how UJT exhibits this
characteristics ?
MULTIPLE CHOICE QUESTIONS
1. As reverse bias voltage is increased, for a diode, the base width at the junction
(a) decreases (b) increases (c) remains same (d) none of these
2. Expression for 'a.' of a BJT in terms of I ,IE' I etc is
pc c
(a)

PNP
Ie
RIJ Vo: -
Vee
+
A-:-+ Is = 400JlA
V
BU
'-----=-=j
)
Fig 5.2 Amplifier circuit. Fig 5.3 Variation of operating point.
From the above circuit,
V CE = Vee - ic x RL .......... ( 5.1 )
V cc is DC Bias Supply. ic is instantaneous value of AC Collector Current.
V CE = Instantaneous Voltage
From the Equation ( 5.1 ), the load line can be drawn. To draw the load line,
Let Ie = 0 then V CE = V CC" This is one point of the load line.
. Vce
Let V CE = 0 then IC = R' This another point of the load line.
L
The value of rc is very small compared to R
L
. Hence· it can be neglected.
t (
Neglecting, ro
270 Electronic Devices and Circuits
So by setting ie = 0, VeE = 0, the load line is drawn. It is a straight line independent of the
transistor parameters. The slope of the line depends on RL the load resistance only. V CC is a fixed
quantity. For a given value ofi
s
' Ie is also fixed when the transistor is being operated In the active
region. Therefore for Is = 200/lA, the quiescent point is QI. For Is = 800 /lA, the operating point
is Q
4
and so on. Now, when the input signal applied between base and emitter is a symmetrical
signal, varying both on positive and negative sides equally, within the range of the transistor
characteristics, then the operating point is chosen to lie in the centre of the output characteristics
(i.e., corresponding to Is:;: 300 /lA), so that the output signal is a true replica of the input without
distortion ( Vee and V ss are bias voltages. Input signal is the a.c signal between Band E ). When
the input signal is symmetrical, the quiescent point is cltosen at tlte centre of tlte load line.
The load line as explained above, if drawn, is called as the dynamic (AC) load line. If the
output impedance in the circuit is reactive, then the load line will be elliptical. Since the voltage and
current are shifted in phase, and the resulting equation will be that of an ellipse. For a resistance
the load line is a straight line.
While drawing a load line, if only the collector resistance Rc is considered and load resistance
RL is infinity ( 00 ), then the load line is drawn with the points [ and V CE ) . It is called as Static
Load Line or DC Load Line
But if the load line is drawn considering a finite load resistance R
L
, and the resistance considered is
RL in parallel with R
e
, it is called AC Load Line or Dynamic Load Line, with AC Input.
There are three types of Biasing Circuits
I. Fixed Bias circuit or Base Bias circuit
2. Collector to Base Bias circuit
3. Self Bias or Emitter Bias circuit (Universal bias or Voltage divider bias
5.2 FIXED BIAS CIRCUIT OR ( BASE BIAS CIRCUIT)
Consider the NPN Transistor Circuit as shown in Fig. 5.4. Re is collector resistance. C
e
is
coupling capacitor. It blocks DC since capacitor is open circuit for D.C. So the output signal will
be pure a.c signal. Rc- limits the current Ie . RB provides the bias voltage for the base. Since this
is NPN transistor, Vee should be positIve, because the collector should be reverse biased.
The drop across R
b
, ( V ce - I x Rs) is negative and it provides positive bias for the base. Since
RL is connected, dynamic load\ne has to be considered. Suppose the output characteristics of the
transistor are as shown in Fig. 5.5.
Suppose Q] is the centre point, and it is chosen as the operating point. But if the input signal
is greater than 40 /lA on the negative side, with QI' the transistor will be cut off stnce for that
swing Is :;: 0, and the transistor is cut off, so another operating point is to be chosen. Choosing the
operating point means, a correct value of Is, Ie' V CEo (if the transistor is in Common Emitter
Configuration) so that we get undistorted output.
Since from the above circuit,
I = VCC-VSE
S RB
I :;: Vec - VCE
e Rc
Transistvr Biasing and Stabilization 271
(
V
o
0
1
C
c
R,
N PN
E
VCE
Fig 5.4 Fixed bias circuit Fig 5.5 Operating point variation
V SE is the base emitter.voltage for a forward biased EB Junction. This will be 0.2V for
Germanium and 0.6 for Silicon. In order to get a large value of Is or the change in the operating
point, Vee or Rs has to be changed. Since once V ce' Rs etc., are fixed, Is is fixed. So the above
circuit is called as Fixed Bias Circuit.
Is is fixed when Rs and Vee are fixed. Because in the expression, for Is' f3 is not appearing.
So once Rs and Vee are fixed, the biasing point is also fixed. Hence the name Fixed Bias Circuit.
5.3 BIAS STABILITY
The circuit shown in Fig. 5.4 is called as a Fixed Bias Circuit, because Is remains constant for
given values of Vee and V ss and Rs. (Is ~ V ss/Rs). So the operating point must also remain
fixed. Suppose the transistor in the circuit is AC 128, and we replace this transistor by another AC
128 transistor .. The characteristics of these two transistors will not be exactly the same. There will
be slight difference. It is because of the limitations in the fabrication technology.
The doping concentration, diffusion length etc., in fabrication cannot be controlled precisely.
If we say impurity concentration is 1 in 10
6
/cm
3
, it need not be the same. The diffusion length will
also be not the same since the temperature inside the furnace during drive in diffusion may vary.
The furnace may not have constant temperature profile along the length of the furnace. Because
of these reasons, transistors of the same type may not have exactly the same characteristics.
Hence in the above circuit, if one AC 128 transistor is replaced by another transistor, the operating
point will change, since the characteristics of the transistor will be slightly differt<,nt.
Hence for the same circuit when one transistor is replaced by the other, operating point
changes. In some cases, because of the change in the operating point, the transistor may be cut
off. In the above circuit, the operating p o i ~ is not fixed, since Is is fixed, and f3 changes. On the
other hand Is should be changed to account for the change in f3 so that the operating point is fixed
or Ie and VeE are fixed.
5.4 THERMAL INSTABILITY·
Change in temperature also adversely effects the operating point. leo doubles for every 10°C rise
in temperature.
. ......... ( 5.2 )
272
leo a.I
s
1=--+--
e I-a I-a
Electronic Devices and Circuits
.......... ( 5.3 )
The collector current causes the collector junction temperature to rise. Hence leo increases ..
So Ie also increases. This in turn, increases the temperature and so leo still increases, and then Ie
So the transistor output characteristics will shift upwards (because Ie increases). Then the operating
point changes. So in certain cases, even if the operating point is fixed in the middle of active region,
because of the change in temperature, the operating point will be shifted to the saturation region.
5.5 STABILITY FACTOR'S' FOR FIXED BIAS CIRCUIT
In the previous circuit, ifIs is fixed, operating point will shift with changes in the values ofIe with
temperature.
leo a.Is
Ie = I-a + I-a
leo gets doubled for every 10°C rise in temperature.
Ie also increases with temperature. So ins is fixed, the operating point will shift. In order to
keep operating point fixed, Ie and VeE should be kept constant. There are two methods to keep
Ie constant.
1. Stabilization Technique:
Here resistive biasing circuits are used to allow 18 to vary to keep Ie relatively constant,
with variation in leO' f3 and V
8E
. i.e., 18 decreases if leo increase, to keep Ie constant.
2. Compensation Technique:
In this method, temperature sensitive devices such as diodes, transistor, thermistors
etc., which provide compensating voltage and currents to maintain the operating
point constant are used.
Stability Factor S is defined as the rate of change of Ie with respect to leo the reverse
saturation current, keeping f3 and V
8E
constant.
ole
s=m-
co p, VBE = K

The value olS should be small. If it is large, it indicates that the circuit is thermally unstable.
Ole ole
, aV
SE
are al"o some times called as Stability Factors.
Expression for Ie'
.......... ( 5.4 )

S - 81
eo p = K, VBE = K
Transistor Biasing and Stabilization 273
Therefore, differentiating Equation (5.4) with respect to Ie' with /3 as constant,
al aI
1 =(1 +/3)x ~ +/3 x ~
ale ale
I = (I + /3) + /3. al s
Sale
leo cannot be measured directly. Ie and Is can be measured directly. So Ie is not
differentiated directly with leo.
(
I - /3. al s) = 1 + /3
ale S
S = 0 implies that Ie is independent ofI
eo
. This is ideal case. S = 5 is a reasonable value for
practical circuit. General Expression for Sis,
1+/3
S = - ~ ' - - - - : -
1- /3(
aI
s)
ale
.......... ( 5.5 )
I = Vee - VSE
s Rs
For a fixed bias circuit, Is is independent of Ie-
als = o.
ale
S=I+/3
If /3 = 40, S = 41. It is a very large value or thermal instability is more. So some circuits are
available which make Ie more independent ofIeo and hence S will be low.
5.6 COLLECTOR TO BASE BIAS CIRCUIT
The circuit shown in Fig. 5.4 is fixed bias circuit, since Is is fixed, if Vee' Re and V ss are fixed.
But an improvement over this circuit is the circuit shown is Fig. 5.6.
~
R"
IB Vee
I :
~
r I
B B
C
B
v,
v
u
,_
1 1 E
Fig 5.6 Collector to base bias circuit.
274
Electronic Devices and Circuits
In this circuit, RB is connected to the collector point C, instead of V ee as in Fig. 5.4. Vi is
input, the capacitor will block DC input and allow only AC signal to the base. Since it is NPN
transistor, collector junction is reverse biased. Re is a resistor in the collector circuit, which controls
the reverse bias voltage of the collector junction with respect to emitter. .
. . VeE = Vee - I Re
1·= Ie + IB
fhis circuit is an improvement over the Fig. 5.4 fixed bias circuit because,
VeE = Vee - (Ie + I
B
) Re
I = Ie + IB
With temperature Ie increases. Since Vee is same, I is same and so IB decreases.
As Ie increases with temperature because of the increases in leo' (Ie + I
B
) Re product
increase. So VeE reduces. That 1l1l':1ns the reverse bias voltage of the colIector junction is reduced
or number of carriers collected I Ill' ,t >l1ector reduces. Hence J
B
also reduces. Therefore, because
of the reduction in the bias CUfl"l'1I1 ' • the increase in Ie will not be as much as it would have been
considering only increase in leo temperature.
We can calculate the stability factor for this circuit. Applying KVL,
V BE - Vee + + Ie) Re + IB . RB = 0
or Vee - V BE =: IB ( Re + R
B
) + Ie . Re
I = Vee - VBE -IeRe
B Re+RB
V BE is the cut in voltage and it is 0.6V for Silicon, and O.2V for Germanium. It is independent
ofIe' So differentiate this expression with respect to leo we get
al O-Re -0 -Re
---1i = ---'=----
ale Re+RB Re+RB
The expression for stability factor S
I+P
S = -1 _-p7( a'--J B---)
8I
e
alB _ -Re
ale Re +Rb
But
l+P
S = ------'-:---
R
1 + /3x e
Re+RB

Therefore, the value of S is less than (1 + /3).
(I + /3) is the maximum value for the fixed bias circuit. Therefore collector bias circuit, is an
improvement over the fixed bias circuit. Now let us consider the effect of /3 on the stability of the
circuit, discussed above.
Transistor Biasing and Stabilization
Vee -leRe - V
BE
I B =
Ie - (1 + P )Ieo
From Equation (5.7), IB = P
Equating (5.8) and (5.9) and transferring Ie to one side
Ie - (1 + p)l
eo
Vee -ieRe - V
BE
P Re +RB
275
.......... ( 5.7 )
.......... ( 5.8 )
.......... ( 5.9 )
Ie (Re + R
B
) - (1 + P)(R
e
+ R
B
) leo = P[Vee - V BE] - PIc Re
Again (P + 1) ::: p.
Ie (Re + R
B
) + P Ie Re = P ([Vee - V BE] + (Re + R
B
) leo (1 + P))
P» 1 = 1 + P ::: P
IdRe + PRe + R
B
] = P[Vee - V BE] + )(Re + R
B
) leo P
Ie [(P + 1) Re + R
B
]::: P [Vee - V
BE
+ (Re + RB)I
o
]
I - P[Vee - V
BE
+ (Re + RB)leo]
e- (pRe+R
B
)
.......... (5.10)
This expression, in this form is derived to get the condition to make Ie independent of p.
Therefore, with the changes in the value of p, with temperature,l
e
changes so the operating point
also changes.
The above circuit, is an improvement over the fixed bias circuit, IB is being decreased, with
increase in Ie (Because leo increases with temperature) to keep operating point constant. Since
Ie is dependent on the P from the above expression, and since P changes with temperature, Ie also
changes and thereby operating point changes. Therefore from the above expression, it is essential
to make Ie insensitive to p, so that the modified circuit works satisfactorily to keep the operating
point fixed. The assumption to be made in the above expression (5.10) is,
P x Re» R
B
·
Therefore, the abOve expression becomes,
I _ P[Vee-VBE+{Re+Rs)IcoJ
e - --'=--------'------'---"'-
PRe
PRe» RB
Ie::: Vee - V
BE
+ (Re + RB)Ieo
Re
So if we choose RB very small compared to PRo the circuit will work satisfactorily. But in
all cases, this may not work out. IfRB value is not so small, then the above assumption is not valid
and hence the circuit won't work satisfactorily.
For a given circuit, to .determine the operating point, draw the load line on the output
characteristics of the given transistor. Load line is drawn as given below:
Vee = Ie RL + VeE
• 276
Electronic Devices and Circuits
If Ie = 0, then, Vee = VeE
Vee
If VeE = 0, thenI
e
= --
RL
With these points, a straight. line is drawn on the output characteristics of the transistor.
Corresponding to a given value ofI
s
' where this line cuts the output characteristic, it is called as
the operating point.
In the collector to base bias circuit we said that the stability is good since, Is is decreased
with increase in Ic to keep operating point fixed. In that circuit, there is feedback from the output
to the input through the resistor Rg- If the signal voltage causes an increase in I
B
, Ie increases. But
V CE decreases. So because of this, the component of base current coming through RB decreases.
Hence the increase in IB is less than what it would have been without feedback. In feedback
circuits the amplification factor will be less than what it would have been without feedback. But
the advantage is stability is improved. Such an amplifier circuit with feedback is called as
Feedback Amplifier Circuit. This type of circuits are discussed in subsequent chapters.
5.7 SELF BIAS OR EMITTER BIAS CIRCUIT
In the fixed bias circuit, since Ie increases with temperature, and IB is fixed, operating point
changes. In collector to base bias circuit, though Ie changes with temperature, IB also changes to
keep operating point fixed. But since P changes with temp. and Ie depends on p, the assumption
:. has to be made that PRe » RB to make Ie independent of p. But if in one circuit, Re is small, the
above assumption will not hold good. [In transformer coupled circuits Re will be small].
So collector to base circuit is as bad as fixed bias circuit. A circuit which can be used even
if there is zero DC resistance in series with the collector terminal, is the Self Bias on Emitter Bias
Circuit. The circuit is as shown. This circuit is also known as Voltage Divider Bias or Universal
Bias Circuit.
+Vcc
i
I (
C
B
Vi
1
N
Fig 5.7 Self Bias circuit. Fig 5.8 Thevenin's equivalent.
The current in the emitter lead causes a voltage drop across Re in such a direction, that it
forward biases the emitter-base junction. It is NPN transistor emitter is n-type. Negative polarity
Transistor Biasing and Stabilization 277
should be at the emitter. If Ie increases due to increase in leo with temperature the current in Rc
als9 increases.
I = Ie + I
B
; V BE = V BN - V EN
As IB increases, Ie also increases.
So current I or emitter current increases. Therefore IRe drop increases. Since the polarity
of this voltage is to reverse bias the E - B junction, IB decrease. Therefore Ie 'will increase less
than it would have been, had there been no self biasing resistor. .
[The voltage drop across Re provides the self bias for the emitter]. Hence stability is good.
With reference to Fig. 5.8, it is NPN transistor. So the conventional current is flowing out of
the transistor, from the emitter. That is why the symbol is with arrow mark pointing outwards. So
emitter point is at positive with respect to N. Hence the drop (IB + Ie) Re has the polarity such as
to reverse bias or to oppose the forward bias voltage V BE junction. RB can be regarded as parallel
combination ofRI and R
2
.
5.8 STABILITY FACTOR'S' FOR SELF BIAS CIRCUIT
S is calculated assuming that no AC signal is impressed and only DC voltages are present. In the
Fig 5.8, the voltage V is the drop across R
2
. Combination acts as a potential divider. So the drop
across R2 which is equal to V, is given by the expression,
V
ee
·R
2
V=R
1
+R
2
'
RB is the effective resistance looking back from the base terminal
RIR2
R =
B RI + R2
Applying Kirchhoff's Voltage Law around the base circuit,
V = IBRB + Re (IB + Ie) + V BE
Assuming V BE to be independent of Ie' differentiating IB with respect to Ie' to get'S' .
I
B
, RB + IB Re = + V - V BE - Ie x Re
IB (RB + Re) = V - V BE - Ie x Re
(v - V
BE
-lcRJ
I = - ' - - - ; - - ~ - - - ' 7 - - = -
B (RB +RJ
alB Re
,al
c
Re + RB
I+P
The expression for stability factor S = I_p(aI
B
)
al
c
278 Electronic Devices and Circuits
Rs .
If R IS very small, S ::: 1.
e
The fixed bias circuit, collector to base bias circuit and Emitter bias circuits provide stabilisation
of Ie against variations in leo' But Ie also varies with V SE and 13. V
BE
decreases at the rate of
2.5 mV,tOC for both Germanium and Silicon transistors. Because, with the increase in T, the
potential barrier is reduced and more number of carriers can move from one side (p) to the other
side (n). 13 also increases side with temperature. Hence Ie changes. .
5.9 STABILITY FACTOR S'
The variation of Ie with V SE is given by stability factors S' .
, al
c
I
S ---
aVSE p = K
where both leo and 13 are considered constant.
V = Is x Rs + VSE + (Is + Ie) Re
Ie = (1 + l3)1eo + 13 . Is
Therefore, From equation ( 5.12 ),
Ie - {I + P )Ieo
Is = P
From equation ( 5.11 ),
V SE = V - IsRs x (Is + Ie) Re
But
Ie - (l + P)l
eo
Is = P
Substituting this value in equation ( 5.13 ),
.......... (5.11)
.......... (5.12)
.......... ( 5.13 )
{
le-O+P)leo } {lc-{l+P)Ico I}R
V SE V - P Rs - P + c e
VSE V_IC.RB+Rb.{I+p)xI +
P P eo
(
1 + P) R x I - x R ..:... I x R
p e eo pee e
V
OE
V -Ie + +R,} + leo )R,}
V
SE
= V+(RS+Re) {1+P}leo- {R
s
+R
e
(I+
P
)}l
e
.......... (5.14)
P P
S'=
aV
SE
aV
SE
=_{Rs +Re{I+ P)}
ale p
Transistor Biasing and Stabilization
or
But
P+l::::P
R B if it small can be neglected
Re
Is' = - S I .. Re
S = 1, implies that the stability of the circuit is very good. Or to a better accuracy,
when RB is small,
Re
R R

Re Re
RB)
Re RB
S = ----;(,...0...1 = 1+ R e
279
If RRB is very large, then S = 1 + p. Now for a fixed R B , S increases with increases p.
e Re
R
If P is small, S is almost independent of p. If a graph is plotted between Sand R B ,then we get,
e
the graph as
i
s
100
10
1.0 RB
Rc
Fig 5.9 Variation oj stability Jactor
280 Electronic Devices and Circuits
The smaller the value of RB the better the stabilisation RB = R) in parallel with R
2
. So
R) and R2 should be small.
The smaller, the value ofR
B
, the better the stabilisation. Because of feedback through Re,
the AC gain is reduced. To prevent this, Re is shunted with a capacitor whose reactance at the
operating frequencies is very small so that AC signal passes through the capacitor only.
, 1 ,
S has the units of n or mhos. S is negative indicate that as V BE decreases, Ie increases.
-S
S'--
- RE
Therefore, ifS is made unity, than S' will also be less. We get stability ofIe with respectto
both V BE and leo. For selfbias circuit, variation ofS"with V BE is also less. S"is approximately
equal to 1.
5.10 STABILITY FACTOR S"
The variation of Ie with respect to is given by the stabil ity factors S".
01

I - +1
e - eo
ale =
op (RB + Ref(1 +
(RB + ReXV - V
BE
)
(RB + Re)+ (1 + pf
"Ole le·
S
S = 8i3 = p{l + p)
5.11 PRACTICAL CONSIDERATIONS
Silicon transistors are superior to Germanium transistor as far as thermal stability is concerned.
The variation in the value ofle for Silicon transistor is 30% for a typical circuit when the temperature
varies from -65°C to + 175°C. But for Germanium transistor, the variation will be 30% when the
Transistor Biasing and Stabilization 281
temperature is varied from -65°C to +75°C. Therefore, the thermal stability is poor for
Germanium transistor. Silicon transistors are never used above 175°C and Germanium transistors
above +75°C. Such high temperature will be encountered in the electronic control of temperature
for furnaces, space applications etc.
5.12 BIAS COMPENSATION
The collector to base bias circuit and self bias circuit are used for stability ofIe with variation in
leo' V SE and 13. But we said that there is feed back from the output to the input. Hence the
amplification will be reduced, even though the stability is improved.
Problem 5.1
An NPN transistor with 13 = 50, is used in Common Emitter Circuit with VeE = 10V and
Re = 2KO. The bias is obtained by connecting a 100 kO resistance from collector to base.
Assume V SE = O. Find +--
( a) The Quiescent Point ( Ie + Is )
( b) The Stability Factor He
Solution
Neglecting leo' Ie = 13 x Is
Then W + 1) Is Re + Is Rs = Vee
(50+ l)x2 x 10
3
+ 100 x 10
3
xI
s
= 10
10
:. Is = 202x 103 = 49.5 /-t
A
Ie = 50 x 49.5 x 10-
6
= 2.475 rnA
51
- ~ - - - - o - = 25.6
1 + 50x2
102
Problem 5.2
B B
E
Ie
~
+
+R ..
r
!
Fig 5.10 For Problem 5.1.
A transistor with 13 = 100 is to be used in Common Emitter Configuration with collector to base
bias. Re = 1 KO Vee =: lOY. Assume V SE = O. Choose Rs so that quiescent collector to emitter
voltage is 4Y. Find Stability Factor.
Solution
Vee - (13 + 1) Is x Re - VeE = 0
101
S = 100x 1 = 41
1+--
68.3
10- 101 x 10
3
Is - 4 = 0
282
Problem 5.3
6
IB = 101xl03 = 59.4Il
A
V CE = IB RB = 4V
4
R = = 67.3K
B 59.4xlO-6
Electronic Devices and Circuits
One NPN transistor is used in the self biasing arrangement. The circuit components values are
V cc = 4.5V, Rc = 1.5K, Rc = 0.27 KO, R2 = 2.7 KQ and Rl = 27K. If ~ = 44. Find the
Stability Factor. Also determine the Quiescent point Q (V CE . Ic).
Solution
,
The Quiescent Point
Vee
27K
+
T
V 2.7K
1
N
Fig 5.11 For Problem 5.3.
Fig 5.12 For Problem 5.3.
Applying KVL over the loop.
+V IB x Rs + V SE + (l + ~ ) RCIB
+0.409 = Is x (2.46 + 45 x 0.27) + 0.2
Transistor Biasing and Stabilization 283
0.409+ 0.2
or
IB = 14.6 = 1.1 rnA
Ie = 13 x IB = 44 x 1.1 = 48.4 rnA
VeE = Vee - Ie x Re - (IB + Ie) Re = 2.4V
If the reduction in gain is not tolerable then compensation techniques are to be used. Thus
both stabilization and compensation technique are used depending upon the requirements.
5.12.1 DIODE COMPENSATION FOR V BE
The circuit shown in Fig. 5.13 employs self biasing
technique for stabilization. In addition a diode is
connected in the emitter circuit and it is forward
biased by the source V DO. Resistance Rd limits the
current through the diode. The diode should be of
the same material as the transistor. The negative
voltage Vo across the diode will have the same
temperature coefficient as V BE. Since the diode is
connected as shown, if V BE decreases with increase
in temperature (V B is cut in voltage), Vo increases
with temperature. Since with temperature, the
mobility of carriers increases, so current through the
diode increases and V 0 increases. Therefore, as V BE
+Vec
decreases, Vo increases, so that Ie will be insensitive Fig 5.13 Diode compensation for V
BE
to variations in V BE.
The decrease in V BE is nullified by corresponding increase in Vo. So the net voltage more
or less remains constant. Thus the diode circuit compensates for the changes in the values of V BE.
5.12.2 DIODE COMPENSATION FOR leo
leo the reverse saturation current of the collector junction increases with temperature. So Ie also
increases with temperature. Therefore, diode compensation should also be given against variation
in leo. A circuit to achieve this is as shown in the Figures 5.14.
+ Vee
NPN
-l,
I
o
1
o
1
Fig 5.14 Diode compensation for I CO'
284,
Electronic Devices and Circuits
The diode D is reverse biased since the voltage V BE which also appears across the diode is
so as to reverse bias the diode. If the diode and the transistor are of the same material, the reverse
saturation current 10 of diode will increase at the same rate as .the transistor collector saturation
current leo'
Vee -VBE
I = - - " = - - ~
RI
Therefore, Vee is measured with respect to ground.
V BE is small compare to Vee'
Vee:::' 15V,
V BE = 0.2 or 0.6Y.
I ~ Vee
- R\
The diode D is reverse biased by an amount 0.2V for Germanium. Current through the
diode is the reverse saturation current 1
0
,
IB = 1-10
But we know that Ie = ( 1+ P) leo + 13 x IB
Substituting the value ofI
B
,
Ie = P x 1- P x 10 + (I + P)Ieo
But P» 1,
P+1:::.13
Ie = P x I - P x 10 + P x leo
-P x 10 and +P x leo are of opposite signs. If the diode and the transistor are of the same
material, the rate of increase of 10 and leo will be the same. Therefore, Ie is essentially is
constant. 10 and leo need not be the same. Only the changes in leo and 10 will be of the same
order. So net value of Ie will remain constant.
5.13 BIASING CIRCUITS FOR LINEAR INTEGRATED CIRCUITS
With the advancements in semiconductor technology, the active components made with this
technology are no more costlier than the passive components like resistors, capacitors etc.
Moreover, incorporating the passive components particularly capacitors in ICs is difficult. With
today's technology, in biasing and stability circuits, transistors and diodes can be used. Hence
certain circuits incorporating transistors in biasing compensation are developed.
One such circuit is shown in Fig. 5.15. Why should transistor be used for compensation
when it is being used as a diode only? It is because of convenience in fabrication technology. In IC
fabrication for a diode, a transistor is being used as a diode. It is one step process and fabrication is easy.
Transistor Q
I
is used as a diode between base and emitter because, collector and base are
shorted. Q
I
is connected as a diode across the emitter base junction of trans i stor Q2' SO effectively
the circuit is same as circuit in Fig. 5.14. Instead of a diode, a transistor is used. Now the collector
current through transistor Q
1
is,
Vee - V
BE
lei = -I
BI
-I
B2
RI
Transistor Biasing and Stabilization
V BE « V ce' and (lBI + I
B2
) « ICI
Vee
ICI = ~ = Constant.
,-------,--- + Vee
I=----v
o
NPN
Fig. 5.15 Biasing circuit for ICs.
285
If transistors QJ and Q
2
are identical, their V BE' (cut in voltage) will be the same. Hence
ICI = IC2 = Constant if RI = Re' These two transistors are being driven by the base voltage V ~ E '
Since, V BE is the same, Ie will not increase. V ce and Rc are also fixed. Hence the stabihty
will be good. Even if the two transistors are not exactly identical, practical experience has shown
that, the stability factor will be around 5. Hence in ICs, such circuits are employed for biasing stability.
5.14 THERMISTOR AND SENSISTOR COMPENSATION
By using temperature sensitive resistive elements rather than diodes or transistors, compensation
can be achieved. Thermistor has negative temperature coefficient i.e., its R decrease
exponentially with temperature.
The circuit is as shown in Fig. 5.16. RT is the thermistor. As temperature increases, RT
decreases. So th current through RT increases. As this current passes through R
e
, there is
voltage drop across Re' Since, it is PNP transistor. Collector is to be reverse biased and hence -
V ce is given. So the potential at the emitter (E) point is negative. Since the emitter is grounded,
this acts as the reverse bias for the emitter. As the drop across Re increases, the emitter reverse
bias increases and hence Ic decreases. Thus the effect of temperature on Ic i.e., due to increase
in temperature, there will be increase in (3, V BE and Ico' The corresponding increase in Ie is
nullified by the increase in the reverse bias voltage at the emitter. Thus compensation is achieved.
V BE = Drop Across R
E
· As current through RE increases, V BE or the forward bias voltage
of E - B junction reduces. Hence Ie reduces. The materials used for thermistors are:
Mixtures of such oxides as NiO (Nickle Oxide), Mn
2
0
3
(Manganese Oxide), (C0
2
)
03 (Cobalt Oxide).
Instead of a thermistor, a temperature sensitive resistor with positive temperature coefficient
like a metal or sensistor can be used. Sensistor is a heavily doped semiconductor material. Its
temperature coefficient will be + 0.7% j0c. When a semiconductor is heavily doped, as temperature
increase, mobility of carriers decrease since there are more number of thermally generated free
carriers (n is large).
286 Electronic Devices and Circuits
-Vee
i
Vi
~ l ___ +o:-__ --'+
Fig 5.16 Thermistor compensation
So temperature compensation can be obtained by placing a sensistor in parallel with R
J
or
Re' Sensistor should be placed in parallel with Re because, to start with, Re will have the same
value of sensistor. So the current will get branched equally. As T increases, Ie increases. Also
the resistance of sensistor increases. Therefore, more current will pass through Re now, since
current takes the path of least resistance. Hence the reverse bias of emitter increases and so Ie
decrease. (Since V BE = Drop across R2 - Drop across Re)' As Drop across Re increases, V BE
decreases hence Ie decreases. So the effect of temperature on Ie is nullified.
5.15 THERMAL RUNAWAY
The max power which a transistor can dissipate (power that can be drawn from the transistor) will
be from mW to a few hundred Watts. The maximum power is limited by the temperature also that
the collector to base junction can withstand. For Silicon transistor the max temperature range is
150°C to 205°C and for Germanium it is 60°C to 100°C. For Germanium it is low since the
conductivity is more for Germanium compared to Silicon. The junction temperature will increase
either-because of temperature increase or because of self heating.
As the junction tempetature rises, collector current increases. So this results in increased
power dissipation. So temperature increases and Ie still further increases. This is known as
Thermal Runaway. This process can become cumulative to damage the transistor. Now
L.\T=TJ-T
A
=8 x Po
where T
J
and T A are the temperature of the junction and ambient. Po is the power dissipated at the
junction in Watts. 8 is the so called Thermal Resistance of the transistor varying from 0.2° C/W
for a high power device to I OOooC/W for a low power device. The condition for thermal stability
is that the rate at which heat is released at the collector junction must not exceed the rate at which
heat can be dissipated. So
apc apo
-- <--
aT] aT]
Transistor Biasing and Stabilization
or
aPe < ~
aT] e
aT]
e = aP
D
287
Heat Sink limit the temperature rise. Heat sink is a good conducting plate which absorbs
heat from the transistor and will have large area for dissipation.
Problem 5.4
Calculate the value of thermal resistance for a transistor having P
e
(max) = 125 mW at free air
temperature of 25°C and T
J
(max) = 150°C. Also find the junction temperature if the collector
dessipation is 75 mW.
(i) T
J
-T
A
=9xP
D
or 150-25=9xI25
(ii) T
J
- 25 = I x 75 ; T
J
= 100°C
Solution
It has been found that, if the operating point is so chosen t h ~ t
Vee
VeE = -2-
then the effect of thermal runaway is not cumulative. But if
V.
V >...s:£.
eE 2
or 9 = I°C/mW
then temperature increases with a increase in Ie- Again with increase in Ie' T increases and hence
the effect is cumulative. As Ie increases, P e increases. But it follows a certain law. If Ie is so
chosen that
Vee
VeE = -2-
rate of increase of P e with Ie is not large and Thermal Runaway problem will not be there.
Problem 5.5
A silicon type 2N780 transistor with f3 = 50, Vee = 10V and Rc = 2500 is connected in collector
to base bias circuit. It is desired that the quiescent point be apporoximately at the middle of the
load line, so IB and Ie are chossen as 0.4 rnA and 21 rnA respectively. Find RB and Calculate'S',
the stability factor.
Solution
V CE = V cc - (lc + I
B
) Rc
IB is very small compared to Ic. So it can be neglected.
.. V CE = V ce - Ie . Rc
= IOV - (21 + 0.4) 250
V
CE
:::: 4.6 V
288
Since it is Silicon transistor,
V
BE
= Vy = 0.6V
VCE - V
BE
IB
4.6-0.6
=IOkn
0.4
[3+1
S = ---'-----=--
1 + [3. Rc
Rc +Rb
Electronic Devices and Circuits
(50+1)
S = 250 = 23
1+50x ( )
250+IOx10
3
S = 23
Problem 5.6
A transistor with [3 = 100 is to be used in Common Emitter Configuration with collector to base
bias. The collector circuit resistance is Re = 1 kn and Vec = 10V. Assume V BE = O.
(a) Choose RB so that the quiescent collector to emitter voltage is 4V
(b) Find the stability factors.
Solution
[3 = 100 ; Rc = I kn
R
=?
b .
S=?
VeE should be 4 V
Vee = 10V
Applyining KVL around the loop,
Vee - (Ic + I
B
) Rc - V CE = o.
V ce is voltage these negative to positive. V CE is voltage drop positive to negative. But Ie
value is not given.
or
But
Ie
- =[3
IB
Ie = [3 x IB
Vec - ([3 + I)IB Rc - VeE = O.
Vee = 10V VeE = 4V Re = I kn [3 = 100 IB = ?
10 - (101) IB x I x 10
3
- 4 = 0
6
IB = 10lxl03 = 59.4 f!A
V
CE
= lB· RB
4 V = 59.4 f!A x RB
Transistor Biasing and Stabilization
4
RB = 59.4x1O-6 = 67.3kO
Therefore, Stability Factor,
Problem 5.7
S= ~ + 1
1 + ~ . Rc
Rc+RB
101
S = =41
1 + 1 00 x 1 x 10
3
10
3
(1+67.3}
S=41
289
Two identical Silicon transistor with ~ = 48, V BE = 0.6V at T = 25° C, V cc = 20.6V, RI = 10K and
Rc = 5K are connected as shown. Find the currents IBi' IB2' ICI and IC2 at T = 25°C
(b) Find IC2 at T = 175° C when ~ = 98 and V
BE
0.22V
Solution
+ Vee
R,
v
C.
~
I
0
vj-_I
!
NPN
Q
1
Fig 5.17 For Problem 5.7.
Since the two transistors are identical, we can assume that IBI = I
B2
.
Vcc -VBE
1= ~ ~ ~ . = .
RI
20.6-0.6
1= 10k =2mA
.. 'YBE = O.6V at 25° C
290 Electronic Devices and Circuits
Since IBI = IB2= I
B
, then I = 21B + ICI = 2IB + J3I
B
or 2mA = (2 + 48)I
B
2
· . IB = SO rnA = 40mA
· . ICI = IC2 = 13 x IB = 48 x 40 x 10-
3
= 1.92mA
13 is same for the two transistors. IB is same. Hence Ic should be the same.
(b) At 17S
o
C, V BE decrease with temperature. Hence V BE at 17S
o
C = 0.22V and
13= 98
VCC-VBE
I = 10k = 2.038mA = (2 + J3)I
B
2.038
or
IB = 2+98 rnA = 2 0 . 3 8 ~ A .
I = I = A X I = 98 x 2038 x 10-6 = 2mA
CI C2 I-' B .
Problem 5.8
Determine the quiescent currents and the collector to emitter voltage for a Ge transistor with
13 = SO in the self biasing arrangements. The circuit component values are V cc = 20V, Rc = 2K, Re
= 0.1 k, RI = lOOk and ~ = Skn. Find the stability factor S.
Solution
Sx20
= 100+S = 0.9S2 V
Fig 5.18 For Problem 5.8.
For Germanium transistor, V BE = 0.2V
· . V = IB x RB + V BE + (lB + IC)Re
Transistor Biasing and Stabilization
Problem 5.9
v = Is x Rs + V SE + + 1) Is Re
= Is [Rs + Re(l ] +V SE
0.952 = I
s
[4.76 x 10
3
+ 51 x 0."1] + 0.2
0.952-0.2
Is = 9.86 X 103 = 76.2 rnA
Ie = x Is = 3.81 rnA
IE = (Is + Ie) = 3.81 rnA
VeE = Vee - leRe - IE Re
=20-3.81 x 2-3.89 x 0.1 = 12V
[ 1 =51 x
51+ 4.76
Re 0.1
291
Design a circuit with Ge transistor in the self biasing arrangement with Vee = 16V and
Re = 1.5K. The quiescent point is chosen to be VeE = 8V and Ie = 4rnA. Stability factor S = 12
is desired :;: 50.
Solution

(d IB)
(d) 1 + 13
5. Voltage divider bias or universal bias circuit is also known as
(a) self bias circuit
(c) collector to base bias circuit
(b) collector bias circuit
(d) Fixed bias circuit
6. By definition, expression for stability factor S" is
(a)
s"
a Ie I
= VBE=K
IeO=K
s"
a
(c)
a Ie
(b)
(d)
s" I =K
a Ie BE' CO
a
s"
- a
312 Electronic Devices and Circuits
7. The units of stability factor S are
(a) ohms (b) constant (c) mhos (d) volt-amperes
8. Compared to silicon BJT, thermal stability for germanium transistors is ..•
(a) good (b) poor (c) same (d) can't be said
9. Sensistors have ...... temperature coefficient
(a) -ve (b) +ve (c) zero (d) None of these
10. The units of thermal resistance
(a) °CIO (b)
OloC
(c)
nloc
(d) °C/O.
In this Chapter,
• A Transistor ( BJT ) can be represented as a group of elements consisting
of Current Source, Voltage S o u t c ~ , Resistor, Conductor, etc. Using this
equivalent circuit, it is easy to aria lyse Transistor Amplifier Circuits.
• The Transistor is represente? in terms of h - Parameters or Hybrid
Parameters.
• The values of these parameters vary with the configuration ofthe transistor
namely Common Emitter ( CE ), Common Base ( CB ) and Common
Collector ( CC ).
• The expressions for Voltage Gain Av' Current Gain AI etc., are derived
using these h - Parameters.
314 Electronic Devices and Circuits
6.1 INTRODUCTION
In chapter, the definitions ofh - parameters, expressions of A
v
' Al etc., in terms ofh - parameters
are gIven.
These are also known as hybrid parameters ( h for hybrid ), so called because, the units of
these parameters are different, Ohms ( n ), mhos ( U ), constants etc. Because the units are
different for different parameters, these are called Hybrid Parameters. In defining these
parameters short circuit and open circuit conditions are used.
In order to analyze different amplifier circuits and compare their merits and demerits,
equivalent circuits of the amplifier must be drawn. These circuits are drawn in terms of the
h - parameters. The transistor equivalent circuits assume that the operating point is chosen correctly
and does not involve biasing resistors.
An amplifier circuit is one which will increase the level or value of the input signal. If the
output voltage (V ) is greater than input voltage (V. ), it is called voltage amplification. Similarly
if P > P., it is amplification. The voltage arhplification factor is represented as A and
° I V
A =V IV.
v 0 I
The other parameters of an amplifier circuit are:
Input resistance R. ( or input impedance Z. )
I I
Output resistance R (or output impedance Z or out admittance Y )
u 0 °
The resistance associated with the amplifier circuit, as seen from the input terminals of the
amplifier, into the circuit is called input resistance ( R.). For a given circuit, R. must be high.
( Ideal value of R. = 00). Because, when the input signJI source V is connected fo the amplifier
circuit, the amplifier circuit will draw current ( I. ) from V s. (1. = V IR.). IfR. is small, I. will be
large, which V s the signal source may not be to supply. So rrlay fall1from the value
( due to loading effect). So R. must be large for a given amplifier CIrcuit.
I
Similarly, the output resistance R of the amplifier must be small. Because, the amplified
signal V 0 must be fully delivered to the looad RL connected to the amplifier. The signal voltage drop
in the amplifier must be as small as possible. So R must be small. Ideally Ro must be zero.
Output resistance is the resistance of the amplifier cft.cuit, as seen through the output terminals
into the amplifier circuit.
6.2 BLACK BOX THEORY
Any four terminal network (two terminals for input and two terminals for output), no matter how
complex it is, can be represented by an equivalent circuit, if there is a connection between input
and output, as shown in the Fig 6.1
II
12
'1'
./
_ A
C _
......
l'
VI Linear circuit
V
2
t
B 0-
t
Fig 6.1 Four terminal network.
Amplifiers 315
If a voltage V 2 is applied at the output terminals and current 12 is allowed to flow into the
circuit, the Thevenins equivalent circuit looking into the input terminals is given by a voltage source
VOl in series with an impedance ZOI. Similarly, if a voltage V I is at the input, the equivalent circuit
can be represented as a voltage source VO) in series with an impedance Z02 so the Thevenins
equivalent circuit reduces as shown in Fig 6.2.
11
ZOI
Z02
12
~ ~
VI
V
2
V V
Fig 6.2 Equivalent circuit of a four terminal network.
But it is convenient to represent the output equivalent circuits by following Nortons Theorem
Vo
as a current generator zo: in parallel with an impedance of Z02. Therefore, the final equivalent
circuit reduces to as shown in Fig 6.3.
II
ZOI
12
~ t
VI V0
2 Z02
V
2
V
Z02
V
Fig 6.3 Equivalent circuit with the current source.
What is the relationship between the Variables VI' II' V
2
, lz' and the constants of the circuits
ZOl' Z02' VOl' V0
2
, etc? The general symbols that are used m these equivalent circuits are
detmedas:
h
n
: Jfvoltage VI is applied to the device, the resulting input current is II then the ratio
V
t represents the resistance. This parameter represents input resistance. Units are
I
Ohms ( (2).
h
tl
: VOl is related to V
2
as VOl = hl2 V
2
. It is called as Reverse voltage gain.
h
21
: The current in the output circuit depends upon the forward current gain and magnitude
of the input current I I. Hence I I and 12 are related as 12 = h21 I I·
h
22
: The output of a device has some output resistance or admittance which is represented
as h
22
. This is output admittance parameter. Units are in mhos ( U )
Therefore, for transistors we use, hll for ZOI or h22 for \/Z0
2
and so on.
316 Electronic Devices and Circuits
The equivalent circuits for a transistor can be represented as shown in Fig 6.4.
II hll 12
Fig 6.4 Equivalent circuit of a transistor.
This is the hybrid equivalent circuit for a transistor. From the equivalent circuit,
VI = hllII + h
l2
V
2
12 = h21 I I + h22 V 2
These h-parameters are constant for a given circuit but they depend upon the type of
Configuration i.e. parameters vary depending upon, whether the circuit is in Common Emitter, or
Common Base or Common Collector Configuration.
So these are represented as h ,h ,hfi and h in Common Emitter Configuration. The
second subscript 'e' indicating Em1tter. Base Configuration, these are
represented as h b' h b' htb and h b' In Common Collector Configuration, the h - parameters are
represented as hi, h 0, h., and h r. The second subscript 'b' and 'c' indicating common base and
IC oc IC rc
common collector.
The general equations are
VI =hlIII +h
12
V
2
12 = h21I1 + h
22
V
2
From the first equation 6.1,
.......... ( 6.1 )
.......... ( 6.2 )
VI
h II = -I when V 2 = 0 ........... ( 6.3 )
J
It has the units of Resistance and V 2 = 0 indicates that output is shorted. Therefore, h JJ. is
the input resistance when output is shorted. It is represented as h. , in Common Emmer
C fi
.
on guratlOn.
h21 = .......... ( 6.4 )
II V2 = 0
This is Forward Current Transfer Ratio or Forward Current Gain, or Forward Short
Circuit Current Gain V 2 = 0 means, output is shorted. Second subscript 'e' in h represents C.E.
Configuration. Ie
h12 = I .......... ( 6.5 )
2 '1=0
It is the ratio of input voltage to the output voltage. Hence it is called as Reverse Voltage
Gain, with Input Open Circuited, since I) = O.
h22 = I .......... ( 6.6 )
2 '1 =0
This parameter is called the Output Conductance when the Input is open circuited.
Amplifiers 317
6.2.1 OUTPUT CONDlJCTANCE
In the audio frequency range, the h - parameters are real numbers ( Integers). But beyond this
frequency range, they become complex.
The reason is that in the audio frequency range, the effect of Capacitance and Inductance
1
are negligible. Capacitive reactance, Xc = 21tfC. If f is small, Xc :: 00 and can be neglected
(Open Circuit). Inductive Reactance XL = 21tfL. If f is small X :: 0 and can be neglected
(Short Circuit). Beyond. audio frequency range, they can't be neg1ected and reactances must
also be considered. Transistor can be represented as a two port active device. It is specified by
two voltages input and output and two currents input and output. The Representation can be
shown as in the Fig 6.5 .
..zS_1 ___ 1_---iITWO Port Active DeVicell-__ O(- __ i2 ___
Fig 6.5 Block diagram.
Current flows only when there is closed path. Input current enters the two port device and
the output current flows through the return path back into the device. Hence the direction ofI2 is
as shown. Now of these 4 quantities (iI' i
2
, vI' v
2
) two can be chosen as independent parameters
and the remaining two can be expressed as variables.
A transformer is a two port device. But V I and V 2 cannot be chosen as independent
VI 12
parameters because V
2
= Hence we cannot choose independent parameters on our choice.
If v 2 and i I are chosen as Independent Variables, then v I and i2 can be expressed in terms
ofv2 and i
l
.
VI = hlli
l
+h
12
v
2
.......... (6.7)
i2 = h
21
i
l
+ h
22
v
2
.......... ( 6.8 )
hll is in n, h\2 is constant, h21 is constant and h22 is in mhos. All these are not the same
dimensionally. Hence these are known as Hybrid Parameters. Suffix 1 indicates input side and
Suffix 2 denotes output side. h-parameters are defined as,
hI I = Input Resistance with Output Shorted. (0)
I V2=0
h
12 v
2 il =0
h21 =
I v2=O
h22 =
2 11=0
Reverse Voltage Gain with Input open circuited
(Dimension Less)
Forward Current Gain with Output Shorted
( Dimension Less ).
Output Conductance with Input Open Circuited ( U )
318 Electronic Devices and Circuits
But, according to IEEE Standards, the notation used is 'i' for 11 ( input) '0' for 22
( output) 'f' for 21 ( Forward Transfer) and 'r' = 12 (Reverse Transfer). The subscripts b, c or
e are used to denote Common Base, Common Collector or Common Emitter Configurations. Thus
for Common Emitter Configuration,
the equation are
-h . +h
v
1
- 11 • v
2 Ie re
.......... ( 6.9 )
i
2
=h
fe
i
1
+ hoe v
2
· .......... (6.10)
If the device has no reactive components, the hybrid parameters are real. But if the device
consists of reactive elements also, h - parameters will be functions of frequency. The four
h - parameters can be used to construct a mathematical model of an Amplifier Circuit. The Circuit
representing the two equations relating v
1
' v
2
' i1 and i2 is shown in the Fig 6.6. h12 v
2
is a Voltage
Source. h21 i2 is a Current Source.
Fig 6.6 h-parameter equivalent circuit
6.3 TRANSISTOR HYBRID MODEL
In drawing the equivalent circuit of a transistor, we assume that the Transistor Parameters are
constant. With the variation in operating points, transistor parameters will also vary. But we assume
that the Quiescent or operating point variation is small.
The advantages of considering h-parameters for transistor are
1. They are easy to measure.
2. Can be obtained from Transistor characteristics.
3. They are real numbers at audio frequencies.
4. Convenient to use in circuit design and analysis.
Manufactures specify a set of h-parameters for transistors.
6.4 TRANSISTOR IN COMMON EMITTER CONFIGURATION
The variables are i
b
, ie' Vb and ve. We can select ib and ve as independent parameters.
v
b
=f
1
(i
b
andv
e
) ........ (6.11)
These are DC quantities Incremental variation (Elementary
Changes) in DC is considered as a.c. quantity. So small/etters v and i
are used..
Therefore, Vb depends upon i
b
, and
i e = f2 ( i
b
, v
e
)
also the Voltage across Collector and Emitter is V CE
........ (6.12)
Amplifiers 319
Making a Taylor Series Expansion of ( 1 ) and ( 2 ) and neglecting Higher Order Terms,
ilv
b
= aa:11 ilib + aaf
1
I ilvC
18 V Vc 1
c b
........ ( 6.13 )
iliC = ilib + a
af
2
1 ilv c
18 V Vc
c Ib
........ ( 6.14 )
ilv ,ilic are A.C. quantities and ( Incremental Variations in D.C. quantities). The quantities
ilv
b
, ili
b
, Xv c and ilic represent small signal (incremental) base and collector voltages and currents.
They are represented as Vb' i
b
, V c' and ic ( A.C. quantities, so small letters are used)
vb = h ib + h Vc ........ ( 6.15 )
Ie re
ic=hfeib+hoevc ........ (6.16)
af aV
b
I
where h = 1 = ai Vc is kept constant ........ ( 6.17 )
Ie U1b B Vc= K
avBI
h --
re - avc
IB= K
........ ( 6.18 )
DC values ofi
b
KI
fe ai B Vc= K
........ ( 6.19 )
ACib=OorvC=O
aic I
h --
oe aVe
Ib=K
( ib is kept constant) ........ ( 6.20 )
Parameters 6.17 to 6.20 represent h-parameters for C.E configuration from equations
( 5 ) and ( 6 ), we can draw the equivalent circuit.
6.5 DETERMINATION OF h-PARAMETERS FROM THE CHARACTERISTICS
OF A TRANSISTOR
The output characteristics of a Transistors in Common Emitter Configuration are as shown in Fig. 6.7.
Q
A
Is = Const.
V CE = Const --. V CE
Fig 6. 7 Transistor output characteristics (To determine hr, ,h graphically)
e oe
320 Electronic Devices and Circuits
h - aiel _ L'1
i
CI
fe - ai
B
- Aib v -0
Ve=K c-
hI£! h can he determined/rom output characteristics. hi£! h,e can he determined
from
Suppose Q is the Quiescent Point around the operating point hfi is to be determined. For
h
fl
, we have to take the ratio of incremental change in ib keeping V From the output
characteristics two values i ,and i are taken corresponCling to ib ib . Then,
CI C2 I 2
h
fe
= (iC2 - i
C1
)/(i
b2
- i
bl
)
h
fc
is also known as Small-Signal Beeta (13) of the transistors. Since we are determining
the incremental values of Current Ratio of i C and i
b
. h
te
is represented as 13·
13 is represented as hfi and is called large signal 13. 13 = i / i
b
, since here no incremental values
are being considered. e c
h = aie I
oe av
e Ib=K
Output characteristics of a transistor are plotted between i and v. Therefore, h is the
slope of the output characteristics, corresponding to a give n value 'bfI
b
at the operating point. For
Common Emitter Configuration, in exactly the same way we have drawn for amplifier system, the
h-parameter equivalent circuit can be drawn. The Transistor in Common Emitter Mode is shown
in Fig. 6.8. The h-parameter equivalent is shown in Fig. 6.9
+-i C
c
C

Fig 6.14 To determine h'J h graphically.
Ie re
From the input characteristics corresponding to two values V c and V c VB and V Bare
V V I 2 I 2
b
2
- b
l
noted then the ratio V _ V gives h
re
·
C
2
C
I
Here we have seen the methods of determining 'h' parameters for Common Emitter
Configuration. The same is true for Common Base and Common Collector Configuration.
6.7 HYBRID PARAMETER VARIATIONS
From the above analysis, it is clear that, when 'Q' the operating point is given, from input and
output characteristics of the given transistors, we can determine the h-parameters. Conversely if
the operating point is changing, the 'h' parameters will also change. Ie changes with temperature.
Hence 'h' parameters of a given transistor also change with temperature because the output and
input characteristics change with temperature . Hence when the manufactures specify typical
h-parameters for a given transistor, they also specify the operating point and temperature. SpecifYing
the operating point is giving the values of If! p and VeE ( DC Values).
h
fe
the small signal current amplification factor is very sensitive to Ic' Its variations is as
shown in Fig 6.15. The variation ofh
fe
with IE and Temperature T are shown in Fig. 6.16.
1
h,.

Fig 6.15 Vqriation of hie with Ie
i
50·C
) IE
Fig 6.16 Variation of hie with IE and T.
Amplifiers 323
We are determining these h parameters from the static D.C characteristics of the transistors
But the' h' parameters are defined with respect to A.C voltages and currents. So AC voltage Vb
may be represented as small changes in D.C. levels of V BE' Hence the above graphical
is correct. We are taking incremental values of DC to represent AC quantities. Typical values of
h-parameters at room temperature, in the three transistor configurations, namely Common Emitter,
Common Collector and Common Base are given in the Table 6.1.
Table 6.1 Comparison of h-parameters in the three configurations (at IE = 1.3 rnA)
Parameter CE CC CB
h I loon 11000 21.60
J
h 2.5 x 10-
4
2.9xI0-4
r
h
f
50 -51 -0.98
ho
24J.lAlV 25J.lAN 0.49J.lAN
6.8 CONVERSION OF PARAMETERS FROM C.B. TO C.E.
The circuit shown in Fig, 6.17 is the h-parameter equivalent circuit in Common Base Configuration.
This circuit is to be redrawn in the Common Emitter Configuration. In Common Emitter
Configuration, Emitter is the common point In C.B, base is common point Hence the above circuit
is to be redrawn so as to get' E' emitter, as the point. Hence invert the voltage source h
rb
V
cb
' From the -ve ofh
rb
V
cb
' we go to htb Ie pomt:

V
es
1
B B
Fig 6.17 h - parameters circuit in C.B. configuration
htb I is as shown. h b is in parallel with hjb I ,V is across h b' I is entering into htb I and h b'
There is nJ connection E and C termmais. !fence the has to be drawn as
the Fig. 6.17. The circuit in Fig. 6.18 is exaGtly the same as Fig. 6.17, But the figure in 6.18 is in
Common Emitter Configuration.
'b ---
B C
B B
thrb v
t
t t
V
BE
hOb
VBCJv
ce
rho
Vee
!
!
E
E
E E
Fig 6.18 (a) Equivalent Circuits Fig. 6.18 (b)
324 Electronic Devices and Circuits
h = ~ b e I = Reverse Voltage Gain
re ce lb."
But Vbe = Vb + V (of redrawn equivalent circuit)
c ce
V
be
+ Vee I
h re = ---"''-v--'''''-
ee 1
8
=0
h = (1 + VbC)
re V
ee 18=0
IfIb = 0, I = -I . Therefore, IB + Ie + IE = 0 for any Transistor. But the current Through h b
is I, (I is in oppoCsite direction. Ie gets brandied through hob). 0
Hence hfb I - I = I
e c
h 1-1=-1
fb e e
or I=hfbl +1.=1 (l +hfb)
e e e
Since hob represents conductance, I = hOb· V be
1= (l + hfb) Ie·
.......... ( 6.25 )
hob V be = (l + hfb) Ie
hob V
bc
or I = --"!<-....!::!:...
e l+hfb
Applying KVL to the output mesh,
h 'b x I + h b V b + Vbe + V = 0
I ere ee
= (hjbhOb)Vbc =-h V +V +V =0
1 + htb rb be be ce
or
Vee hjbhob + (1- h
rb
)(1 + h
tb
)
Hence
h = 1 + Vbc = hjbhob - (1 + htb)hrb
re Vee hjbhob + (1- hrb)(l + h
tb
)
But
hjbhob - (1 + htb )hrb
h
rb
« l,andhobh
'b
«(l+hfb)= (l+h
tb
).
h'bhob _ h
h
re
== 1 + h rb,
tb
Hence, .......... ( 6.26 )
Amplifiers 315
h = Vbe
Ie Ib
v"''"''
Ifwe connect terminals C and B together, we get the circuit as (the same redrawn circuit in
Common Base Configuration with C and B shorted) is shown in Fig. 6.18. Applying KVL around
the I mesh on the left hand side,
V be + h
lb
Ie - h
rb
V cb = O.
hOb I = + h b Vb - Vb
I ere e
Vbc=Vbe
C b
·· h . I (I - hrb)y
om mmg t ese two equatIOns, = - h be
e ,b
Applying KCL to node B,
1+1 +h I -h V =0
be tbe obbe
J-h
or I = (1 + h ) __ r_b V + h V
b tb h
lb
be ob be
Hence
then
6.9 MEASUREMENT OF h-PARAMETERS
........... ( 6.27 )
In the circuit shown in Fig. 6.19 to determine the h-parameter, PNP transistors are taken into
account. The operating point is chosen by suitably adjusting R
2
, VEE and V ce' there by adjusting
the values of Ic and V CB' The h-parameters depend upon the frequency ot operation. So the
.....--------,--------i I 0 0
Fig 6.19 Circuit for determining h - parameters.
326 Electronic Devices and Circuits
signal generator is adjusted to be 1 KHz, at which frequency normally all the transistor h parameters
are specified by the manufactures. 1 KHz is in the audio frequency range. So the h-parameters
are assumed to be real at audio frequency. The tank Circuit impedance will be very high at this
frequency, of the order of500kO. This is very high compared to the input resistance of the transistor.
The tank circuit is tuned to the same frequency of the signal generator (1KHz). This will prevent
any stray pick up. The stray signal will pass through the LC tank circuit, since it offers least
impedance. So spurious signal will not affect the net voltages and currents of the transistor, while
determining the h-parameter.
The name, tank circuit, is used for LC Parallel circuit, because energy is· stored in these
elements, as water is stored in a tank.
Vb I . I - Vs
h
ie
= ,
vc
Since the input resistance of the transistor is small compared to the Tank Circuit Resistance,
all the current Ib flows into the transistors alone. Rs is negligible compared to ( 1 MO ).
I = Vs
b RI
Now hand hfj are determined when V = 0 or the collector is shorted to ground. But if the
collector is dfrectly connected to the ground,1 will be different and V CE will be different. The
operating point will change h parameters. on the operating pomt h - parameters differ.
Hence AC short circuit is to be done. C:z. block D.C. Hence the D.C values of V CE' I will not
change at all. But the A.C. Voltage of V is = 0 or the collector is at ground potentta{for A.c.
voltages. So this is called AC short circuit.
Now
h = JCRI
fe Vs
where V 0 is the voltage across Rr
hand h are defined when Ih = 0 or input is open circuited. Therefore, the tank circuit
impedahece is large. The input Sloe can be regarded as open circuit. The signal generator is
connected on the output side.(as shown by the dotted lines).
h = Vbl = Vb
re Vc Ib=O Vc
Amplifiers 327
6.10 GENERAL AMPLIFIER CHARACTERISTICS
An amplifier generally produces an enlarged version of an input signal. The amplifier may be
thought of as a black box, that has two input terminals; and two output terminals for the connection
of the load and a means of supplying power to the amplifier. One of the input and output terminals
will be common for the input and output ( Fig. 6.20 ).
DC Source
AMPLIFIER
Fig 6.20 Block schematic of amplifier.
The signal to be amplified may be AC or DC. Input signal is a low level voltage such as the
one obtained from a tape head, or a transducer like thermocouple, pressure gauge etc. The output
load can be a loud speaker in an audio amplifier, a motor in a servo amplifier or a relay in control
applications. In ~ n y case, the output of the amplifier is an enlarged version of the input to amplify
means to increase the size. So in the case of electrical signals, if it is voltage or power amplifications,
there is some additional energy being gained by the output signal. Energy can be neither created
nor destroyed. So this additional energy is being gained from the D.C. bias supply. So without Vee
or DC Bias Voltage, (in the case of transistor amplifier circuit), the circuit will not work.
The notations that are used in the case of amplifier circuit are ( Fig. 6.21 )
Vee
AMPLIFIER
Fig 6.21 Block schematic.
V Open circuit signal voltage
s
I Source or signal current.
s
R Internal resistance of the source in n.
s
R Amplifier input resistance.
I
V Amplifier input voltage
I
V = Amplifier output voltage
o
I = Amplifier output current
o
R Load Resistance.
o
328
Current Gain
Voltage Gain
. I. - io peak to peak
A = 10 15 - •
I 15 peak to peak
_ V 1 V = Vo peak to peak
Av - 0 I VI peak to peak
p V·
Power Gain A = -.JL= ~ =A A
P p. V.is v· I·
I I
Electronic Devices and Circuits
An amplifier mayor may not exhibit both voltage and current gains, but in general, it
exhibits power gain.
6.10.1 AMPLIFIER INPUT RESISTANCE R.
J
The amplifier circuit presents a load R to the source which can be 50 n to few thousand n for
transistor circuits. I
The input voltage to the amplifier will be reduced from V by an amount that depends on
both Rand R. (Fig. 6.22 ) s
S I
V=iR
5 S
V.
I
v
s
1
Fig 6.22 Input side equivalent circuit for an amplifier.
R·
V=Vx I
I S R. +R
I S
1
=Vx--
sIRs
+-
R
j
IfR.» R the input voltage to the amplifier is same as that of V . or ifR is small then also
V. = V ushally It = 50 n. From the above equation, we can determir:e the input resistance of a
g i ~ e n :mplifier ci}cuit by noting the open circuit voltage ofthe voltage source V and then voltage
V. after connections to the amplifier. If R of the voltage source is known, R. can be calculated.
I S I
6.10.2 AMPLIFIER OUTPUT RESISTANCE Ro
R. is used to denote the load resistance. So Ro is the output resistance of the amplifier circuit.
(Fig. 6.23).
The amplifier can be represented as a voltage source in series with internal resistance of
Ro in parallel with 1\ is the external load resistance K is the amplification factor.
Amplifiers 329
__ R.."'-L_ = KV x ---
I 1 Ro
+-
V =Ky'
o I
RL
Output voltage V can be made large by increasing the value ofRL But the output current
will be small and so the °current gain will fall. Hence power gain will also be less. In order to get
maximum power gain, R, should be matching with R . It should be complex conjugate of R
according to Maximum P'ower Transfer Theorem. For °this purpose, matching transformers a r ~
used (:: = ~ : ) in the output stage of an amplifier. Similarly matching transformers are also
used to couple one stage of an amp lifer to the second stage, if the input resistance of the second
stage amplifier is very small.
i
v
o
KV .
•
1
Fig 6.23 Output side equivalent circuit.
Conversion Efficiency :
An amplifier circuit draws energy from the D.C. source and amplifies the A.C. signal. The D.C.
supply provides the major portion of the extra energy.
. A.C. signal power delivered to the load I 00 ~
ConversIOn TJ =. .. x =:: p .
D.C. tnput power to the actIve devIce DC
This is also known as collector circuit efficiency in the case of transistor and plate circuit
efficiency in the case of tube amplifier circuits.
Problem 6.1
A single stage amplifier employing one active device, is powered by a 9V battery which has a
current drain of20 rnA. Ifthe load voltage is 3V at 12 m A, determine the conversion TI.
Po == Volo == 3x 12x 10-
3
==36mW
P
oc
==Vocloc ==9x20xI0-
3
==180mW
n=:: 36 xI00=20%
.• 180 .
330 Electronic Devices and Circuits
6.11 ANALYSIS OF TRANSISTOR AMPLIFIER CIRCUIT USING h-PARAMETERS
To form transistor amplifier configuration, we connect a load impedance ZL and a signal source as
shoWn in Fig. 6.24.
R
1 II 12 2 s
+
t +t
,1. IL
VI
Transistor V
2
ZL
,1. -,1.
l'
2'
Fig 6.24 Amplifier circuit.
V is the signal source, R is the resistance of the signal source, and ZL is the load impedance.
can be connected insC.E, C.B. and C.C. Configuration. To analyse these circuits i.e. to
determine the current gain AI' Voltage gain A , input impedance, output impedance etc, we can use
the h parameters. So the equivalent circuit fo; the above transistor amplifier circuit in general form
without indicating C.E, C.B, or C.C. Configuration, can be denoted as in the Fig. 6.25.
Rs 1 II hi 2
,1.-
l'
Fig 6.25 . h-parameter equipvalent circuit (general repersentation).
Current Amplification,
I
A _-1.,.
I - II'
But 1 = - 1
L 2
A
I I'
1
Negative sign is because 12 is always represented as flowing into the current source.
F or a transistor PNP or NPN type, if it is flowing out of the transistor, 12 is represented as - 1
2
,
Now voltage across Z2 is taken with 2 as +ve and 21 as -ve The directIons of IL are as shown
and IL = -1
2
,
From the above circuit we have,
12=hfll+hoV2
But V
2
= IL ZL = - 1
2
, ZL (Since IL =-1
2
)
Amplifiers
or
A =_ -hf
I II 1 + hoZL
Negative sign indicates that I) and 12 are out of phase by 180°.
6.11.1 INPUT IMPEDANCE: Z.
331
.......... ( 6.28 )
Input impedance of any circuie is the impedance we measure looking back into the amplifier
circuit. Now amplifier terminals are 1 and 11. R is the resistance of the signal source. So to
determine the output Z ofthe amplification alone, Sit need not be considered.

G ....... ---'-------+----'---....... G
Fig 6.64 e.G configuration.
Vo = 1D x (rd II R
L
)
Amplifiers
Zo = R2 II rd :::: RL
10 = gm . V gs = gm . VI
V 1
Z=_I=-
I ID gm
This is the input impedance ofthe device.
The circuit input impedance is, _1_11 Rs.
gm
6.18 COMPARISON OF FET AND BJT CHARACTERISTICS
From construction point of view triode is similar to BJT (Transistor).
Cathode E ; B ~ Grid; Anode ~ collector.
But from characteristics point of view pentode is similar to a BJT.
369
1. In FET and pentode current is due to one type of carriers only. In pentode it is only due
to electrons. But in BJT it is due to both electron and holes.
2. In FET pinch off occurs and current remains constant. But no such channel closing in
Transistor and Pentode.
3. As T increases, leo increases, so I in a BJT increases. But in FET, as temperature
increases, '11' decreases. So 10 decreases. In pentode, also as temperature increases,
I increases. But it is less sensitive compared to a transistor or FET.
Pentode Transistor
Ip
/
Ie
/
t t
-"T -..T
Fig 6.65
4. As Is decreases Ie decreases in a transistor.
As V GS decreases, 10 decreases in FET.
As V control grid increases Ip decreases in pentode.
t
FET
ID
~
t
-..T
._-----V GS=O
_------1
~ - - - - - - 2
-"V
os
Fig 6.66
370 Electronic Devices and Circuits
5. FET and pentode are voltage dependent devices. Transistor is current dependent.
6. ro ofFET is very large. MO, rc of transistor is less (in kO).
7. Breakdown occurs in FET and BJT for small voltages 50, 60V etc. But in pentode it is
much higher.
8.
BJT
Saturation (it IS very narrow)
t I: A";,, reg;"
Cutoff region
~ V C E
JFET
Ohimic region (it is more)
'D I: ~ Pentode or
t . : Saturation region
Fig 6.67
6.19 R. C. COUPLED AMPLIFIER
The circuit is as shown in Fig. 6.68 In the A.C. equivalent circuit, the collector output of the
transistor (BJT) is coupled to the output point V 0 through a resistor (R) and capacitor (C). RL
and Co are the coupling elements. So it is called R-C coupled amplifier .
.,....----1 f--r----O Vo
I
Output
Fig 6.68 R.C coupled amplifier circuit.
Rl' Rand RC' RE are biasing resistors. It is a self bias circuit. The operating point
'Q' is locate; at the centre of the active region of the transistor characteristics. V CE will be
about V c ~ , typically 5 to 7 V. V
BE
must be 0.5 V for silicon transistor. IB will be about 100
IlA and Ic 5 mAo These are the proper bias conditions. Rs is the source resistance of the input
A.C source. If an external resistance is connected on the mput side, it is to limit the input A.C.
voltage so that the BJT is not damaged due to excess input current. Ca..is the blocking capacitor.
It blocks D.C components present in the A.C source, so that the D.c biasing provided is not
Amplifiers 371
changed. C
c
is the coupling capacitor, to couple the amplified A.C signal at the collector of the
transistor to the output point V 0 RE is emitter resistor to provide biasing, so that V 0 - V EO
= Y BE.= 0.5 to 0.6 V for silicon transistor. Rc is collector resistor to limit the collecPor current
from V CC' to protect the transistor. C
E
is emitter bypass capacitor. It by passes A.C signal,
so that A.C signal does not pass through R
E
, C
E
is chosen such that X
CE
= ){o R
E
, at the
lower cut off frequency f1' X
CE
will be any way smaller than R
E
.
The A.C equivalent circuit is as shown in Fig. 6.69.
hie -.I
b
.-----,----,----'\

t
Fig 6.69 A.C equivalent circuit.
Frequency Response is a plot between frequency and voltage gain. Voltage gain is measured
in decibels.
I P
I decibel = 10 bel. = 10 log pO
,
y2
P =_0
o R
y2
P.= -'
1 R
yg;, [Y J
'R 0
Gain = 10 log = 20 log V db
Y, ' ,
IR
As the frequency of the input signal varies, gain of the amplifier also varies, because
of the variation of the reactance of capacitors with frequency. At high frequencies, capacitance
associated with BJT also must be considered.
f1 or fL is called Lower cut off frequency or Lower half power point or Lower 3db
frequency (f2 - f
1
) is called Bandwidth.
I
f1 and f2 are chosen such that the gain at these frequencies is .J2 or 0.707 of the maximum
gain value. At f1 and f2' output power is half of the maximum value. Because at these frequencies
( )
2 I
y y2 y2
= ;;; R = --'l!... P = --'l!... So f and fare
-v 2 2 R max R ) 2
I
. Y S . y2
vo tage IS --'l!... 0 output power IS -
.J2 R
called as half power points. At f) and f2' Po = P ;ax . In decibels it is 20 log ::: 3db. fo f1
and f2 are also called as 3 db points.
372 Electronic Devices and Circuits
FREQUENCY RESPONSE
The graph of voltage gain versus frequency is as shown in Fig.6.70.
"
L
z
7z ---, -- -- -- -- -----
B.W
I

Fig 6.70 Frequency response.
I
Initially as frequency increases, capacitance reactance due to C
b
, u
Cb
= 27tfC
b
decreases.
So the drop across it also decreases and hence net input at the base of transistor increases and
V
so V 0 increases. Hence Ay = VO also increases.
I
In the mid frequency range, XCb is almost a short circuit, since its reactance decreases
I
with increasing frequency. In the low frequency range, XCc = 27tfC
c
is open circuit (in the
A.C equivalent circuit).
In the mid frequency range, the capacitance reactances of both Cb, and C are negligible.
So the net change with frequency, remains constant as shown in the figure. C
As the frequency is further increased, capacitance reactance due to C
c
?ecreases. Hence
V 0 decreases because, V 0 is taken across C . So as frequency increases, -Y,n decreases and
hence Ay decreases. So rue shape of the cufve is as shown in the Fig. 6.71J.
-hfeRL
Expression Ay (M.F) = R h
S + ie
I
PHASE RESPONSE
Phase shift between Vo and Vi varies as shown in Fig. 6.71.

----+/
Fig 6.71 Phase response.
Amplifiers 373
6.20 CONCEPT OF fa' fp AND fT
In order to obtain some idea of a transistor's high frequency capability and what transistor to
choose for a given application, we examine how transistor's CE short-circuit forward-current gain
varies with frequency. When RL = 0, the approximate high-frequency equivalent circuit is drawn below.
B
r
bb
, ,
C
B
I
+
~ \ .
V
b'e
Cb'e +C
b
, c
~ gm Vb'e
1
E E
Fig 6.72 Approximate high-frequency circuit.
The current gain,
IL -gm Vb'e
AI= T= I.
I I
Ii = V b'e gb'e + V b'e [jro (C
be
+ Cb'c)]
A - -gm Vb'e _ -gm
I - Vb'e [gb'e + jro(Cb'e + Cb,c) ] - gb'e + jro(Cb'e + C b'c)
.: gb'e = gm
hfe
A = -gmhfe
I gm + j21tf(C
b
'e + Cb'c)h fe
-hre -hre
= [(. '2 f(C C )h fe ] = 1+j21tf(Cb'e+
C
b'c)
J + J 1t b'e + b'c
gm
[
.• r: h
fe
]
. b'e
gm
- h ~
A = e
I 1+ { ~ )
1
Where f. = ------
p 21tTb'e(Cb'e +Cb'c)
h
fe
= CE small signal short-circuit current gain.
374 Electronic Devices and Circuits
P Cut-off Frequency 'fP' :
The '[3' cut-off frequency f ~ , also referred as f
hfe
, is the CE short-circuit small-signal forward-
current transfer ratio cutoff frequency. It is the frequency at which a transistors CE short-circuit
current gain drops 0.707 from its value at low frequency. Hence, f ~ represents the maximum
attainable bandwidth for current gain of a CE amplifier with a given transistor.
a Cut-off Frequency 'fa' :
A transistor used in the CB connection has a much higher 3-dB frequency. The expression for the
current gain by considering approximate high frequency circuit ofthe CB with output shorted is
given by
Where
fa is the alpha cutoff frequency at which the CB short-circuit small-signal forward current
transfer ratio drops 0.707 from its 1KHZ value.
Gain Bandwidth Product :
Although fa and fb are useful indications of the high-frequency capability ofa transistor, are even
more important characteristics is fr.
The fr is defined as the frequency at which the short-circuit CE current gain has a magnitude
of unity. The 'fr' is lies in between f ~ and fa.
h
IA 1= fe
I )l+(f/fpt
at
f ~ f" IA,I ~ I, ( ~ ; ] ' » I,
we obtain fr / f ~ '" .hre
I fr '" httJp I
:. fr is the product of the low frequency gain, h
re
, and CE bandwidth fW fr is also the
product of the low frequency gain, h
th
, and the CB bandwidth, fa. i.e fr = h$ . fa. The value off
r
ranges from 1 MHz for audio transistors upto 1 GHZ for high frequency transistors.
Amplifiers 375
The following Fig. 6.73 indicates how short-circuit current gains for CE and CB connections
vary with frequency.
AI
100
(CE)
50 hfel-----__
0707 hfc
Gam Bandwidth
1 ____________ ___
o a Cut off
(CB)
I
lU
0
7-, --:",:-;---.l...;---'-;--..!-"":"I::;:O,--r
l
Frequency In liZ
* v *
0.1
fT fa
Fig 6.73
SUMMARY
• Hybrid Parameters are h - parameters, so called because the units of the four
parameters are different. These are used to represent the equivalent circuit of
a transistor.
• The h - parameters and their typical values, in Common Emitter Configuration
are,
•
•
•
•
h. = 500 n· h = 1.5 x 10-4· h = 6 lin· h = 250·
Ie 're 'oe r-' fe '
The second subscript denotes the configuration. similarly band c represent
Common Base and Common Collector Configurations.
The Transistor Amplifier Circuit Analysis can be done, using h-parameters,
replacing the transistor by h-parameter equivalent.
Expressions for Voltage Gain A
v
' Current Gain AI' Input Resistance Ri' Output
Resistance R etc., can be derived in terms of h-parameters.
o
'J,b' is called the base spread resistance. It is the resistance between the
fiction base terminal Bf and the outside base terminal B.
376
1.
2.
3.
4.
5.
6.
•
Electronic Devices and Circuits
C.E. Amplifier circuit characteristics are:
(a) Low to moderate Ri (300 0 - 5 kO)
(b) Moderately high Ro (100 kO - 800 kO)
(c) Large current amplification
(d) Large voltage amplification
(e) Large power gain
(f) 180
0
phase shift between Vi and Vo.
• For C.B. amplifier:
Low R
i
. High R
o
' Ai < 1, Ay > 1, No phase inversion, Moderate Ap
• ForC.C. amplifier,
High R
i
, Low R
o
' Large AI' Av < 1, Low Ap
• In Darlington pair circuit, the two BJTs are in C.C. configuration. The charac-
teristics are :
High Ri' Large AI' Very low R
o
' Ay < 1
• In CASCODE amplifier configuration, one BJT in C.E. configuration is in se-
ries with another BJT amplifier in C.B. configuration. Its characteristics are,
Very large A
y
, Large AI and thigh R
o
.
• In the case ofR-C coupled amplifier, the output is coupled to the load through
Rand C, hence the name. The lower cut-off frequency 1; is also called lower
3-db point or lower half power point. The upper cut-off frequency h is also
called upper 3-db point or upper half power frequency. In the low frequency
range and high frequency range, the gain decreases. In the mid frequency
range, the gain remains constant, if2 - fi) is called Band width.
OBJECTIVE TYPE QUESTIONS
The units of h-parameters are ............................. .
h-parameters are named as hybrid parameters because .............................. .
The general equations governing h-parameters are
VI······························
12 = ............................. .
The parameter h is defined as h = ............................. .
re re
h-parameters are valid in the .............................. frequency range.
Typical values of h-parameters in Common Emitter Configuration are
7. The units of the parameter hare ............................. .
rc
8. Conversion Efficiency of an amplifier circuit is ............................. .
9. Expression for cun·ent gain AI in terms ofh
fe
and h
re
are AI = ........................ .
Amplifiers
10. In Common Collector Configuration, the values of h - ............................. .
rc -
11. In the case of transistor in Common Emitter Configuration, as I\. increases,
R ................... .
I
12. Current Gain A
J
of BJT in Common Emitter Configuration is high when I\. is
13. Power Gain of Common Emitter Transistor amplifier is ................... .
14. Current Gain AI in Common Base Configuration is ................... .
15. Among the three transistor amplifier configurations, large output resistance is
in .................... configuration.
16. Highest current gain, under identical conditions is obtained in ................... .
transistor amplifier configuraiton.
17. C.C Configuration is also known as .................... circuit.
18. Interms of h
fe
, current gain in Darlington Pair circuit is approximately
19. The disadvantage of Darlington pair circuit is ................... .
20. Compared to Common Emitter Configuration R of Darlington piar circuit is
I
21. In CASCODE amplifier, the transistors are in .................... configuration.
22. The salient featuers of CASCODE Amplifier are ................... .
ESSAY TYPE QUESTIONS
1. Write the general equations in terms of h-parameters for a BJT in Common
Base Amplifiers configuration and define the h-parameters.
2. Convert the h-parameters in Common Base Configuration to Common Emitter
Configuration, deriving the equations.
3. Compare the transistor ( BJT ) amplifiers circuits in the three configurations
with the help of h-parameters values.
4. Draw the h-parameter equivalent circuits for Transistor amplifiers in the three
configurations.
5. With the help of necessary equations, discuss the variations of Av' AI' R, R
o
' Ap
with Rs and RL in Common Emitter Configuration.
6. Discuss the Transistor Amplifier characteristics in Common Base Configura-
tion and their variation with Rs and RL with the help of equations.
7. Compare the characteristics of Transistor Amplifiers in the three configurations.
8. Draw the circuit for Darlington piar and derive the expressions for AJ' Ay, R, and Ro'
9. Draw the circuit for CASCODEAmplifier. Explain its working, obtaining over-
all values of the circuit for hI' hI' ho and hr'
377
378 Electronic Devices and Circuits
MULTIPLE CHOICE QUESTIONS
1. The h-parameters have
(a) Same units for all parameters (b) different units for all parameters
(c) do not have units (d) dimension less
2. The expression defining the h-parameter h22 is,
(a) h22 = ::1 II =0 (b) h22= : ~ I II =0
(c) h22 = I2.I1
(d) h22 = ~ J II = 0
3. h - Parameters are valid over a ...... frequency range
(a) R.F. (b) ForDe only
(c) Audio frequency range (d) upto I MHz
4. By definition the expression for his,
oe
(a)
a Ie I
alB IE=K
(b)
a Ie I
aVe IB=k
(c)
a Vel
a Ie IB=k
(d)
a IB I
a VB le=k
5. Typical value of htb is
(a) -0.98 (b) -101 (c) + 2.3 (d) 26.6 n
6. The expression for h. interms of htb and h.
b
is, h. -
Ie I Ie -
htb ~ ~ ~
(a) hIe"::::: -- (b) I+h (c) I+h (d) I h
I-h tb tb - tb
7. The ratio of A ~ signal power delivered to the load to the DC input power to the
active device as a percentage is called
(a) conversion (b) Rectification 11 (c) power 11 (d) utilisation factor
8. The general expression for AI interms of h
f
ho and ZL is ... AI =
ho h
f
ho h
f
(a) l+hfZ
L
(b) l+hoZL (c) I-hfZ
L
(d) l-hoZL
9. Expression for Avs in terms of A , R , ZL' Z., Z is .....
S S I 0
AI Aj.ZL ZL
(b) (c) (d)
Rs + Z, Rs - Z, Rs + Z, Rs + Z,
(a)
10.
In the case of BJT, the resistance between fictious base terminal B' and
outside
base terminal B is, called
(a) Base resistance
(b) Base drive resistance
(c) Base spread resistance
(d) Base fictious resistance --------------------------------- ----
Amplifiers 379
11. In large signal analysis of amplifiers ....
(a) The swing of the input signal is over a wide range around the operating point.
(b) Operating point swimgs over large range
(c) stability factor is large
(d) power dissipation is large
12. The units of the parameter hoe ...
(a) (b)
°
(c) constant (d) V-A
13. Typical value of hre is
(a) 10-4 VIA (b)
10-
4
(c) 10-
6
AN (d) 10
4
14. CASCODE transistor amplifier configuration consists of
(a) CE-CE amplifier stages (b) CE-CC stages
(c) CE-CB stages (d) CB-CC stages
15. For CASCODE amplifier, on the input side, the amplifier stage is in ....
(a) C.B configuration (b) c.c. configuration
(c) emitor follower (d) C.E. configuration
16. CASCODE amplifier characteristics are
(a) Low voltage gain, large current are
(b) Low voltage gain, low current gain
(c) large voltage gain, large current gain, high output resistance
(d) Large current gain, Large voltage gain Low putput resistance
17. Characteristics of Darlington circuit are typical values
(a) Ri=MO, AI= 10k, Ro=MO,Av= 100
(b) Ri = MO, AI = 10 k, Ro = 100, Av < 1
(c) Ri = 100, AI = 1, Ro = 1.00, Av > 1
(d) Ri=MO,A,= I,Ro= 100,Av< 1
18. The disadvantage of Darlington pair circuit is ...
(a) low current gain (b) low output resistance
(c) leakage current is more (d) high input resistance
19. Compared to C.c. configuration, Darlington pair circuit has
(a) low current gain (b) low voltage gain
( c ) large current gain (d) large p output resistance
20. In the case of Darlington circuit, AI is approximately
(a) 2 h
fe
(b) 4 h
fe
(c) h
fe
380 Electronic Devices .•• d Circuits
Specifications of Amplifiers
Parameters Typical Values
1. Type of Amplifier Audio / Typical Values / Power / RF / Video
2. Frequency Range 15 Hzs - 100 KHzs
3. Output Power 200mW
4. Voltage gain 20db
5. Current gain 100
6. Power gain 9
7. Input impedance 10 kO 5 pf
8. Output impedance 500n 1 pf
9. Band width 100 KHzs
, ' .
. , . ~ .
, . ~
~ : ~ ' .
.. : " , , \ ~ ..
. : ~ "
. ~ 'r'
382 Electronic Devices and Circuits
7.1 FEEDBACK AMPLIFIERS
Any system whether it is electrical, mechanical, hydraulic or pneumatic may be considered to have
at least one input and one output. If the system is to perform smoothly, we must be able to measure
or control output. If the input is 10m V, gain of the amplifier is 100, output will be I V. If the input
deviates to 9m V or 11 m V, output will be 0.9 V or 1.1 V. So there is no control over the output. But
by introducing feedback between the output and input, there can be control over the output. If the
input is increased, it can be made to increase by having a link between the output and input. By
providing feedback, the input can be made to depend on output.
One example is, the temperature of a furnace. Suppose, inside the furnance, the temperature
should be limited to 1000°C. If power is supplied on, continuously, the furnace may get over heated.
Therefore we must have a thermocouple, which can measure the temperature. When the output of
the thermocouple, reaches a value corresponding to 1000oC, a relay should operate which will
switch off the power supply to the furnace. Then after sometime, the temperature may come down
below 1000oc. Then again another relay should operate, to switch on the power. Thus the
thermocouple and relay system provides the feedback between input and output.
Another example is traffic light. If the timings of red, green and yellow lights are fixed, on
one side of the road even if very few vehicles are there, the green light will be on for sometime.
On the other hand, if the traffic is very heavy on the other side of the road, still if the green lamp
glows for the same period, the traffic will not be cleared. So in the ideal case, the timings of the red
and green lamps must be proportional to the traffic on the road. If a traffic policeman is placed, he
provides the feedback.
Another example is, our human mind and eyes. Ifwe go to a library, our eyes will search
for the book which we need and indicates to the mind. We take the book, which we need. If our
eyes are closed, we can't choose the book we need. So eyes will provide the feedback.
Basic definitions
Ideally an amplifier should reproduce the input signal, with change in magnitude and with or without
change in phase. But some ofthe short comings of the amplifier circuit are
I. Change in the value of the gain due to variation in supplying voltage, temperature or due
to components.
2. Distortion in wave-form due to non linearities in the operating characters of the
amplifying device.
3. The amplifier may introduce noise (undesired signals)
The above drawbacks can be minimizing if we introduce feedback
7.2 CLASSIFICATION OF AMPLIFIERS
Amplifiers can be classified broadly as,
I. Voltage amplifiers.
2. Current amplifiers.
3. Transconductance amplifiers.
4. amplifiers.
This classification is with respect to the input and output impedances relative to the load
and source impedances.
Feedback Amplifiers 383
7.2.1 VOLTAGE AMPLIFIER
This circuit is a 2-port network and it represents an amplifier (see in Fig 7.1). Suppose
R,» Rs, drop across Rs is very small.
R,
Fig 7.1 Equivalent circuit of voltage amplifiers.
V,::::V
s
·
Similarly, ifR
L
» Ro, Vo::::Av· VI"
ButV,- Vs.
Vo::::Av· Vs·
:. Output voltage is proportional to input voltage.
The constant of proportionality Ay doesn't depend on the impedances. (Source or load).
Such a circuit is called as Voltage Amplifier.
Therefore, for ideal voltage amplifier
R, = 00.
Ro= o.
Vo
A
y
=-
VI
with RL = 00.
Ay represents the open circuit voltage gain. For ideal voltage amplifier, output voltage is
proportional to input voltage and the constant of proportionality is independent of Rs or R
L
.
7.2.2 CURRENT AMPLIFIER
An ideal current amplifier is one which gives output current proportional to input current and
the proportionality factor is independent ofR
s
and R
L
.
The equivalent circuit of current amplifier is shown in Fig.7.2.
',---+ r - - - - - - - - - - - - ,
R,
Fig 7.2 Current amplifier.
384
For ideal Current Amplifier,
Rj=O
Ro= 00.
If
Ro= 00,
IL = 10 = Aj Ij = Aj Is.
IL .
AI = T with RL = O.
I
Electronic Devices and Circuits
.. AI represents the short circuit current amplification.
7.2.3 TRANSCONDUCTANCE AMPLIFIER
Ideal Transconductance amplifier supplies output current which is proportional to input voitage
independently ofthe magnitude ofR
s
and R
L
.
Ideal Transconductance amplifier will have
Rj = 00.
Ro= 00.
In the equivalent circuit, on the input side, it is the Thevenins' equivalent circuit. A voltage
source comes in series with resistance. On the output side, it is Norton's equivalent circuit with a
current source in parallel with resistance. (See Fig. 7.3).
~ ~ = ~
,...---J\/\
+
Fig 7.3 Equivalent circuit 0/ Transconductance amplifier.
7.2.4 TRANS RESISTANCE AMPLIFIER
It gives output voltage Vo proportional to Is, independent of Rs a. R
L
. For ideal amplifiers
Rj =0, Ro=O
Equivalent circuit
R,
Fig 7.4 Trans resistance amplifier.
Norton equivalent circuit on the R
j
input side.
Thevenins' equivalent circuit on the output side.
Feedback Amplifiers 385
7.3 FEEDBACK CONCEPT
A sampling network samples the output voltage or current and this signal is applied to the input
through a feedback two port network. The block diagram representation is as shown in Fig. 7.5.
GENERALIZED BLOCK SCHEMATIC
-"'1, -...1
Signal
.,.
omparatio[
.,. +
Basic
t +
Sampling
Sou ree Vs or
v.
amplifier
V
Network
"
mixer
i - A .. -
+
Ir
t + Feedback
vr
network
-t -
p
Fig 7.5 Block diagram of feedback network.
Signal Source
It can be a voltage source V s or a current source Is
FEEDBACK NETWORK
-... =
0 L
t +
~
Vo
.. ' - .f»
It is a passive two port network. It may contain resistors, capacitors or inductors. But usually a
resistance is used as the feedback element. Here the output current is sampled and feedback. The
feedback network is connected in series with the output. This is called as Current Sampling or
Loop Sampling.
A voltage feedback is distinguished in this way from current feedback. For voltage feedback,
the feedback element (resistor) will be in parallel with the output. For current feedback the element
will be in series.
COMPARATOR OR MIXER NETWORK
This is usually a differential amplifier. It has two inputs and gives a single output which is the
difference of the two inputs.
V = Output voltage of the basic amplifier before sampling [see the block
diagram of feedback]
V, = Input voltage to the basic amplifier
Av = Voltage amplification = V N,
AI = Current amplification = III,
G
M
= Transconductance of
basic amplifier = IN i
R
M
= Transresistance = VII,.
386 Electronic Devices and Circuits
All these four quantities, A
v
, Af, G
M
and RM represent the transfer gains (Though G
M
and
RM are not actually gains) of the basic amplifier without feedback. So the symbol 'A' is used to
represent these quantities.
Ar is used to represent the ratio of the output to the input with feedback. This is called as the
transfer gain of the amplifier with feedback.
AVf= Vo
Vs
10
A
1f
= -
Is
10
G
Mf
= -
Vs
Vo
R
Mf
= -
Is
Feedback amplifiers are classified as shown below.
Positive
(regenerati ve)
Series
Feedback
I
I
Voltage
I
A = voltage gain = Vo
V,
N
I.
egatIve
(degenerative)
I
Shunt Series
I
Current
I
V
Feedback Factor, 13 = V
f
(This 13 is different from 13 used in 8JTs )
o
(I3)(A)(- 1) = -I3A
Loop gain / Return Ratio
Shunt
- I is due to phase-shift of 180° between input and output in Common Emitter Amplifier. Since
Sin(1800) = - I.
Return difference 0 = I-loop gain negative sign is because it is the difference
0= 1-(-[3A)
10= 1+ [3·A.1
Types of Feedback
How to determine the type of feedback? Whether current or voltage? If the feedback signal is
proportional to voltage, it is Voltage Feedback.
If the feedback signal is proportional to current, it is Current Feedback.
Feedback Amplifiers 387
Conditions to be satisfied
I. Input signal is transmitted to the output through amplifier A and not through feedback
network 13.
2. The feedback signal is transmitted to the input through feedback network and not through
amplifier.
3. The reverse transmission factor 13 is independent of Rs and R
L
.
7.4 TYPES OF FEEDBACK
FeedbacK means a portion of the output of the amplifier circuit is sent back or given back or
feedback at the input terminals. By this mechanism the characteristics of the amplifier circuit can
be changed. Hence feedback is employed in circuits.
There are two types of feedback.
I. Positive Feedback
2. Negative Feedback
Negative feedback is also called as degenerative feedback. Because in negative feedback,
the feedback signal opposes the input signal. So it is called as degenerative feedback. But there are
many advantages with negative feedback.
Advantages of Negative Feedback
I. Input impedance can be increased.
2. Output impedance can be decreased.
3. Transfer gain Afcan be stabilized against variations in It-parameter of the transistor with
temperature etc.
i.e. stability is improved.
4. Bandwidth is increased.
5. Linearity of operation is improved.
6. Dist0l1ion is reduced.
7. Noise reduces.
7.5 EFFECT OF NEGATIVE FEEDBACK ON TRANSFER GAIN
Fig 7. 6 Block scltematic for negative feedback.
Vo
Ay=y-
I
A'v= Vo/V,'
VI' = V,-I3V
o
388 Electronic Devices and Circuits
v 0 = Ay (V i - ~ V 0)
Vo = Ay. VI - Ay . ~ Vo
V 0 (1 + ~ Ay) = Ay V I
Vo A,
Ayf = V = 1 + n. A
I I-' v
7.5.1 REDliCTION IN GAIN
(for positive feedback)
Ay = Voltage gain without feedback. (Open loop gain).
If the feedback is negative, ~ is negative.
Av
A 'v = 1 - {- p.A J
Av
For negative feedback A'v = ---
I+PA
v
Denominator is > I.
:. There is reduction in gain.
7.5.2 INCREASE IN BANDWIDTH
If, fH is upper cutoff frequency.
fL is lower cut off frequency.
f is any frequency.
Expression for Ay (voltage gain at any frequency f) is,
Av{mid)
A = - - - - - ' - - ' - - ~
y 1 + j.L
fH
Ay (mid) = Mid frequency gain
Av (mid)
Ay = 1 + PAy for negative feedback.
Substituting equation (1) in (2) for Av,
Ay{mid)/1 + j.L
A = fH
y 1 + p[ Aymid 1
1 + jf / fH
(for negative Feedback, ~ is V
e
)
..... (l)
..... (2)
Feedback Amplifiers
Simplifying,
Ay (mid) [1 + jL]
fH + j.f fH
1+ j.f [Ay mid]
A =
y
Ay (mid)
fH + j.f +
, AyCmid)11 +
Ay = jf
1 + -.--------"'--,--------.
fH[l +
Equation (3) can be written as,
Where
A'=
y
(mid)
1+
fH
AvCmid)
Ay' (mId) = 1 + (mid)
and fH' = fH (1 + Ay (mId»
is negative for negative feedback,fH' > fH.
:. Negative feedback, increases bandwidth.
Similarly,
or
A '= Ay
y 1+ j
f
7.5.3 REDUCTION IN DISTORTION
Suppose, the amplifier, in addition to voltage amplification is also producing distortion D.
Vo = Ay . V, + D.
Where, Vi = V,' - Vo (for negative feedback)
Vi' = Vi - Vo and is negative for negative feedback,
V
o
' =Ay
389
..... (3)
390
Electronic Devices and Circuits
Vo [l+Ay Py]=V,',Ay+D
V = V:,Av + 0
o (1 + Ay.f3v) (1 + f3vAv)
For negative feedback, f3 is negative therefore denominator is> 1
The distortion in the output is reduced.
o
--- is<D
1 + PvAv
The physical explanation is, suppose the input is pure sinusoidal wave. There is distortion in
the output as shown, in Fig. 7.7.
V,
t
---+ t
(a) Input signal
Fig 7.7
(b) Output with distortion
(without negative feedback)
Now, if a part of the distorted output is fed back to the input, so as to oppose the input, the
feedback signal and input will be out of phase,
So the new input Vi will have distortion introduced init, because of mixing of distorted
Vfwith pure VI' So the distortion in the output will be reduced because, the distortion intro-
duced in the input cancels the distortion produced by the amplifier. Because these two distor-
tions are out of phase, The feedback signal cancels the distortion produced by the amplifier,
Therefore these two are out of phase.
7.5.4 FEEDBACK TO IMPROVE SENSITIVITY
Suppose an amplifier of gain Al is required. Build an amplifier of gain A2 = DA, in which D is
large. Feedback is now introduced to divide the gain by D. Sensitivity is improved by the same
factor D, because both gain and instability are divided by D, The stability will be improved by the
same factor.
7.5.5 FREQUENCY DISTORTION
A
A=--
f l+PA
If the feedback J'U:twork does not contain reactive elements, The overall gain is not a function
offrequency. So frequency duration is less, If f3 depends upon frequency, with negative feedback,
Q factor will be high.
Feedback Amplifiers
7.5.6 BAND WIDTH
It increases with negative Feedback ( as shown in Fig.7.2)
/11 </1,12
1
> /2
BW' = (j;'- BW = (/2 - /1 )
, 1 'f A ' BW
f.. = 12= 2 (l+f-.. .A
m
) BW >
BW = /2 - fa :::'/2
(BW)f = -/1-
(BW)f = (1 + I3Am) BW
wIthout feedback
Gam - - - - -- - - - - -__ --------....Jt
i
wIth negatIve feedback
,
.Ii
Fig 7.8 Frequency Response with and without negative feedback.
A
AmplifIer
13
Feedback Network
Fig 7.9 Feedback network.
7.5.7 SENSITIVITY OF TRANSISTOR GAIN
391
Due to aging, temperature effect etc., on circuit capacitance, transistor or FET, stability of the
amplifier will be affected
Fractional change in amplification with feedback divided by the fractional change without
feedback is called Sensitivity of Transistor.
dAf
Af
Id:1
392 Electronic Devices and Circuits
Differentiating,
dA
f
I IdAI
Af = 11+J3AI A
I
: . Sensitivity, = I + ~ A
Reciprocal of sensitivity is Densitivityl D = (1 + I3A).\
7.5.8 REDUCTION OF NONLINEAR DISTORTION
Suppose the input signal contains second hannonic and its value is B2 before feedback. Because of
feedback. B2f appears at the output. So positive I3B2 f is fed to the input. It is amplified to - AI3 B2 f .
.. Output with two tenns B2 - AI3 B2 f.
B2 - AI3 B2 f = B
2f
·
B2
B2f = I + J3A B2f < B2
or
So it is reduced.
7.5.9 REDUCTION OF NoIsE
Let N be noise constant without feedback and NF with feedback. NF is fed to the input and its value is
J3N
F
. It is amplified to - J3AN
F
.
as
NF = N - J3ANF
NF (1 + J3A) = N
N
N =--
F 1 + J3A
NF < N. Noise is reduced with negative feedback.
7.6 TRANSFER GAIN WITH FEEDBACK
Consider the generalized feedback amplifier shown in Fig. 7.10.
Comparator
mixer ~ +
Difference signal
Xs
Input signal
Feedback signal
,---------,
Xd= X, Basic
Feedback
Network 'p'
Amplifier A
Fig 7.10
Output signal
X =AX
o I
RL
External
load
The basic amplifier shown may be a voltage amplifier or current amplifier, transconductance
or transresistance amplifier.
Feedback Amplifiers
The four different types of feedback amplifiers are,
1. Voltage series feedback.
2. Voltage shunt feedback.
3. Current series feedback.
4. Current shunt feedback.
Xs = Input signal.
Xo = Output signal.
~ = feedback signal.
Xd = Difference signal. [Difference between the input signal and the
feedback signal].
f3 = Reverse transmission factor or feedback factor = X
f
/ Xo.
393
The feedback network can be a simple resistor. The mixer can be a difference amplifier.
The output of the mixer is the difference between the input signal and the feedback signal.
Xd = Xs - X
f
= XI·
Xd is also called as error or comparison signal.
~ = . &
Xo
It is often a positive or negative real number.
I
This f3 should not be confused with the f3 of a transistor f
B
A
_Xo_Xo
----
XI Xd
Transfer gain
X=X
I d
Gain with feedback Af = Xo
Xs
But X =X -X =X
d s f I
Xf
f3= -
Xo
From (2), Xd = Xs - Xr
Substitute ( 2) in (1).
Xo
A= ----"--
Xs -X
f
Dividing Numerator and Denominator by X
s
'
X a / ~ X a / ~
A= -x;- Xc Xa
1-- 1--.-.
~ X a ~
..... (1 )
..... (2)
..... (3)
394 Electronic Devices and Circuits
f = gam WIt feedback.
A = transfer gain without feedback.
If I A ~ < IAI the feedback is called as negative or degenerative, feedback
If I A ~ > IAI the feedback is called as positive or regenerative, feedback
7.6.1 LOOP GAIN
In the block diagram of the feedback amplifier, the signal Xd which is the output of the comparator
passes through the amplifier with gain A. So it is multiplied by A. Then it passes through a feedback
network and hence gets multiplied by 13, and in the comparator it gets multiplied by-I. In the
process we have started from the input and after passing through the amplifier and feedback
network, completed the loop. So the total product AI3.
In order that seriesfeedback is most effective, the circuit should be driven from a constant
voltage source whose internal resistance Rs is small compared to R, of the amplifier. IfRs is very
large, compared with R"V, will be modified not by V
f
but because of the drop across Rs itself.
So effect of V
f
will not be there. Therefore for series feedback, the voltage source should have
less resistance. (See Fig. 7. II ).
R
s
Amplifier
Fig 7.11 Series feedback.
In order thatshuntfeedback is most effective, the amplifier should be driven from a constant
current source whose resistance Rs is very high ( Rs » RJ
R
s
Fig 7.12 Current shunt feedback.
Feedback Amplifiers 395
If the resistance of the source is very small, the feedback current will pass through the
source and not through R. So the change in I will be nominal. Therefore the source resistance
, ,
should be large and hence a current source should be used.
If the feedback current is same as the output current, then it is series derived feedback.
When the feedback is shunt derived, output voltage is simultaneously present across RL and
across the input to the feedback. So in this case V f is proportional to Vo'
V
,
Fig 7.13 Series derived feedback.
An amplifier with shunt derived negative Feedback increases the output resistance. When
the feedback is series derived, I remains constant, so R increase.
o 0
Similarly if the feedback signal is shuntfed it reduces the input resistance.
Fig. 7.14 (a) Shunt derived feedback.
If it is series fed, it increases the output resistance. Therefore 10 remains constant, even
through V 0 increase, so ~ increases.
Return Ratio
f3A = Product of feedback factor [3 and amplification factor A is called as Return Ratio.
Return Difference (D)
The difference between unity (1) and return ratio is called as Return difference.
D= 1-(-f3A)= l+f3A.
Amount of feedback introduced is expressed in decibels
= N log (A') N = 20 log Af = 20 log 1_1_1
A A 1 +AP
If feedback is negative N will be negative because A I < A.
396 Electronic Devices and Circuits
7.7 CLASSIFACTION OF FEEDBACK AMPLIFIERS
There are four types of feedback,
1. Voltage series feedback.
2. Voltage shunt feedback.
3. Current shunt feedback.
4. Current series feedback.
Voltage Series Feedback
Feedback signal is taken across RL.proportional to Vo. So it is voltage feedback. V
f
is coming in
series with VI So it is Voltage series feedback.(See Fig.7.l4).
V
I
Fig 7.14 Schematic for voltage series feedback.
V
V
I
I
(a) Voltage shunt Feedback (b) Current Shunt Feedback
V
I
r-----....J I\.
(c) Current Series Feedback
Fig 7.15
Improvement of Stability with Feedback
Stability means, the stability of the voltage gain. The voltage gain must have a stable value, with
frequency. Let the change in Av is represented by S.
Feedback Amplifiers
= Vo (for negative feedback)
Vs
, Ay
A = ---'--
y 1 + Av
. ... dAy
Dlfferentlatmg wIth respect to Ay, A = ( )2
y
. 'd' b b h'd f dA'y 1
Dlvl tng y Avon ot Sl es, ° -d- = ( )2
Ay
1 1 1
Av dAy = (1
But
A' and dA,'y = S' dAy =S
v = 1 + Ay Ay
dA' 1
S=-_v =---.,,-
dAy (1 f
For negative Feedback, 13 is negative:. denominator> I.
S' < S
i.e., variation in A
y
' or % change in A, is less with -negative feedback.
:. Stability is good.
7.8 EFFECT OF FEEDBACK ON INPUT RESISTANCE
7.8.1 INPlJT RESISTANCE WITH SHlJNT FEEDBACK
With feedback
If = 13, 1
0
, [-: = :: 1
I=I+AI
SIt-' 0
397
398 Electronic Devices and Circuits
, V,
R =
, I, +p,A,I,
V,
R = ----,---'------,-
, I,(I+p,A,)
, R,
R =
, (I + p, A,)
Input resistance decreases with shunt feedback.
If the feedback signal is taken across R
L
, it is a Vo or so it is Voltage feedback.
If the feedback signal is taken in series with the output tenninals, feedback signal is proportional
to 1
0
, So it is current feedback.
If the feedback signal is in series with the input, it is seriesfeedback.
If the feedback signal is in shunt with the input, it is shuntfeedback.
EXPRESSION FOR R WITH CURRENT SHUNT FEEDBACK
I
AI = Shunt circuit current gain of the BJT
AI = practical current gain 10111
I
,
R AI
1 1 1
Fig. 7.16 Current shunt feedback.
AI represents the Shunt circuit current gain taking Rs into accounts,
Is = II + If = II + f3 10
and
VI R,
R - - --'--
If - (l+PA])I, - l+p.A]
for shunt feedback the input resistance decreases.
+
Feedback Amplifiers
7.8.2 INPUT IMPEDANCE WITH SERIES FEEDBACK
V,' = V, + 0 (in general case).
For negative feedback, is negative.
V,'
I, -
In general, R. increases,
Vi __ ---. __ V-"--II
I l
A
I
J
L
J
I
V' = V + Vo
Fig 7.17 Feedback network.
, I
V' == V + A.A. V
lit-' 1
V,' = V,
But V' = I R
, I I
V'
-1-' = Input Z seen by the source = R, (I
,
R
,f
= R,
EXPRESSION FOR R WITH VOLTAGE SERIES FEEDBACK
I
399
In this circuit Ay represents the open circuit voltage gain taking Rs into account. (see Fig.7.18).
or
Let,
R
I Fig. 7.18 Voltage series feedback.
Vl=VS-Vf
Vs=V\+Vf
= R\f = V s/I\.
V
f
= 1\ R\ + V
f
== 1\ R\ +
Ay V,R
L
Vo = R R = (Ay 1\ R\)
0+ L
+
400 Electronic Devices and Circuits
Vo AvRL
A = -=--'---"'----
V VI Ro + RL
V S = Ii RI + 13 . V 0
R
If I VI
AV = voltage gain, without feedback.
for series feedback input resistance increases.
7.9 EFFECT OF NEGATIVE FEEDBACK ON Ro
Voltage feedback (series or shunt) Ro decreases.
Current feedback (series or shunt) Ro increases.
Series feedback (voltage or current) R increases.
,
Shunt feedback (voltage or current) R, decreases.
OUTPUT RESISTANCE
Negative feedback tends to decrease the input resistances. Feeding the voltage back to the input
in a degenerative manner is to cause lesser increase in Vo. Hence the output voltage tends to
remain constant as RL changes because output resistance with feedback ROf« RL"
Negative feedback, which samples the output current will tend to hold the output current
constant. Hence an output current source is created (R
Of
» R
L
). So this type of connection
increases output resistance.
For voltage sampling ROf< Ro. For current feedback ROf> Ro.
7.9.1 VOLTAGE SERIES FEEDBACK
Expression for ROf looking into output terminals with RL disconnected,
+
R
,
Fig. 7.19
RO is determined by impressing voltage 'V' at the output terminals or messing '1', with input
Rof terminals.-shorted.
To find Rot' remove external signal (set Vs = 0, or Is = 0)
Let RL = 00.
Impress a voltage V across the output terminals and calculate the current I delivered by V.
Then, ROf = VII.
Feedback Amplifiers
Because with
Hence,
Vo = output voltage.
VI = -l3
v
V.,=O,VI =-v
f
=-l3
v
Vo
Rof= I =
This expression is excluding R
L
. [fwe consider RL also Rof is in paraIlel with R
L
.
~ ( - ) -xR
1 + PAv L
Ro/ = ---'---'---
Ro
- ~ - + RL
l+PA
v
Substitute the 1 value of Rof
7.9.2 CllRRENT SHllNT FEEDBACK
~ I
I
I
s
R
I
AI
I I
Fig. 7.20 Block schematic for current shunt feedback amplifier.
10= -I
Ai = Shunt circuit current gain
Al = Practical current gain ( I I ~ J
V
1= R -A( I(
o
With, Is = 0, I( = - If = -131 = + 131,
V V
I = - - 13 A I or I (I + 13 A ) = -
Ro ( (R
o
V
Rof = [= Ro (I + 13 A)
A( = short circuit current gain.
Roc' includes RL as part of the amplifier.
401
402 Electronic Devices and Circuits

So if XI and X
2
are capacitive, X3 should be inductive and vice versa.
442 Electronic Devices and Circuits
If XI and X
2
are capacitors, the circuit is called Colpitts Oscillator (Fig. 8.15(a))
If XI and X
2
are inductors, the circuit is called Hartely Oscillators (Fig. 8.15(b))
L
C
(a) Colpitts oscillator (b) Hartely oscillator circuit
Fig. 8.15
8.7.1 FOR HARTELY OSCILLATOR
Condition for oscillations, XI + X
2
+ X3 = O.
where
jroL
I
+ jroL2 + -. _1 - = 0
Jro C
1
or roL + roL = --
I 2 roC
1 1
ro
2
(LI + L
2
) = C ; = ro
8.7.2 FOR COLPITTS OSCILLATOR
j -j
X =- -; X
2
=-; X3 =jroL.
I roC
I
roC
2
XI + X
2
+ X3 = 0;
---=-.L - ---=-L + jro L = 0 or
roC
I
roC
2
Oscillators 443
1 1
roL= -- +--
roC) roC
2
roL or ro = (F, 1
2.kl
where
c = C
1
C
2
T C)+C-)
8.8 WIEN BRIDGE OSCILLATOR
In this circuit, a balanced bridge is used as the feedback network. The active element is an
operational amplifier. It employs lead-lag Network. Frequency fo can be varied in the ratio of 10 :
1 compared to 3 : 1 in other oscillator circuits.
External voltage V 0' is applied betweeen 3 and 4, as shown in Fig.8.16.
V
1
+
i i
Fig. 8.16 Wien bridge oscillator circuit.
To find loop gain, - (-sign because phase-shift feedback)
I
C
R
Lead Network:
II
YIv
Same I is passing through C and R.
V
So I leads V.
C
Lag Network :
I
1
I lags with respect to V.
Fig. 8.17
444 Electronic Devices and Circuits
Frequency variation of 10: I is possible in Wein Bridge compared to 3 : 1 in other oscillator
circuits.
Loop gain
Vo=AyVI;VI is(V
2
-V
I
); VI=V
f
·
_ Vo _ Av·VI _ ~ A
- y.----y:--- .
o 0
VI and V
2
are auxiliary voltages.V
I
= V
2
- VI·
V
-13 = Vi, .: Yr =\; :. A=A,
o
c ~
v
,
Fig. 8.18
8.9 EXPRESSION FORI
V
,
Fig. 8.19 Wien Bridge oscillator circuit.
..... (1)
Oscillators
RJR = (R2 R+ .R2 ) (1 + jwCR)
JWC
R2 · R + =
[(R R
2
) (jCOC) + R
2
] R2 jcoCR - co
2
C2 R2 R2 = R
J
RjcoC
(R
J
R
2
) (jCOC) + R2 + R2 jeoCR - C0
2
c
2
R2R2 = R
J
R jcoC
Equating imaginary parts,
R
J
R2 coC + R2 coCR = R
J
ReoC
Equating real parts,
A = 1 + R = eo
2
C2 R2 R
R2 C
2
' 2 2
or I
or
-R2
--=roCRR
2
roCR
,,2 ; orl 1
So the minimum gain of the
amplifier must be 3.
445
This is the frequency at which the circuit oscillates. Continuous variation of frequency is
accomplished by using the capacitors 'C'.
8.10 THERMISTOR
Conductivity of Germanium and Si increases with temperature. A semiconductor when used in this
way, taking advantage of this property is called a Thermistor. Ex : NiO, Mn
2
0
3
etc. These are
used for temperature compensation in oscillator circuits.
8.11 SENSISTOR
A heavily doped semiconductor can exhibit a positive temperature coefficient of resistance because
under heavy doping, semiconductor acquires the properties of a metal. So R increases because
mobility decreases. Such a device is called Sensistor. These are also used for temperature
compensation like thermistors.
8.12 AMPLITUDE STABILIZATION
The amplitude of oscillations can be stabilized by replacing R2 with a senistor. If is fixed, as A
increases, amplitude of oscillations increases. If a senistor is introduced, as its 'R' changes with
temperature, it changes so that is always constant, (when A changes).
446 Electronic Devices and Circuits
By equating the imaginary part to zero we get f = ~ where K = RR
C
21t RC 6+4K
Forward Current Gain I h
fe
= 4k + 23 + ¥·I
Practically RC phase-shift oscillators can be used from several hertz to sev.eral hundred
kilohertzs. In the mega hertz range, tuned LC circuits are more advantageous. Frequency of
oscillators can be changed by changing Rand C. Amplitude of oscillations will not vary if any C is
varied, because Xc varies, but the imaginary part will be zero. Phase-shift oscillator is operated in
class A in order to keep distortion minimum.
8.13 ApPLICATIONS
Elevation levelling systems, Burglar detection: frequency of oscillators change as a result oflocal
disturbance. The change in frequency causes further electronic action or alarm signal.
8.14 RESONANT CIRCUIT OSCILLATORS
Initial transient due to switching will initiate electrical signals. These are feedback to the transistor
as input. The input is amplified and obtained as output. Oscillations are sustained. Output is
coupled to the base oftransistor through L\. LL\ is a transformer. R represents the resistance in
series with the winding in order to account for the losses in the transformer shown in Fig. 8.20. If
1
Or' is very small, then at ro = ~ the Z is purely resistive. Then voltage drop across inductor is
"LC
180
0
out of phase with the applied input voltage to the FET. If the direction of winding of the
secondary connected to the gate is such that it introduces another 180
0
phase-shift, the total
phase-shift is zero.
R
e
C"
Fig. 8.20 Resonant circuit oscillator.
The ratio of the amplitude of secondary to the primary voltage is MIL where M is
the inductance.
. V
f
M
I.e., V = L = ~
o
Oscillators
Voltage gain Av =-Il
f3 A
v
=-1
-1lf3 = -1;
1
Il= -
13
L
Il= -
M
Il= LIM
If we consider resistance 'r' also, 00
2
= L ~ (l + ~ )
I ~ ~ ~ L · I
8.15 CRYSTAL OSCILLATORS
447
When certain solid materials are deformed, they generate within them, an electric charge. This
effect is reversible in that, if a charge is applied, the material will mechanically deform in response.
This is called Piezoelectric effect.
Naturally available materials: 1. Quartz 2. Rochelle salt.
Synthetic materials: 1. Lithium sulphate 2. Ammonium-di-hydrogen phosphate, PZT
(Lead Zirconate Titanate), BaTi0
3
(Barium Titanate).
If the crystal is properly mounted, deformations take place within the crystal, and an electro
mechanical system is formed which will vibrate when properly excited. The resonant frequency and
Q depend upon crystal dimensions etc. With these, frequencies from few KHz to MHz and Q in the
range from 1 OOOs to 100,000 can be obtained. Since Q is high, and for Quartz, the characteristics are
extremely stable, with respect to time, temperature etc., very stable oscillators can be designed. The
frequency stability will be ± 0.00 I %. It is same as ± 10 parts per million (l0 ppm).
The electrical equivalent circuit of a crystal is as shown in Fig.8.21. L,C,R are analogous to
mass, spring constant, and viscous damping factor of the mechanical system.
a ,..------a
c:::J
C
lb
b
Crystal Electrical Model
Fig. 8.21
448 Electronic Devices and Circuits
Values for a 90 kHz crystal are L = 137 H, C = 0.023pF; R = 15kO corresponding to
Q = 5,500. The dimension of a crystal will be 30 x 4 x 1.5 mm. C' is the electostatic capacitance
between electrodes with the crystal as a dielectric. C' = 3.5 pF and is larger than C.
When the crystal slabs are cut in proper directions, with regard to the crystal axis, a
potential difference exists between the faces ofthe crystal slab when pressure is brought to bear
on them. And if the slab is placed in an electrostatic field, the slab undergoes deformation. (Fig.8.22).
If the electric field is an alternating one, with a frequency which sets the slab into mechanical
resonance, the slab will physically vibrate vigorously. Such a crystal can be employed to maintain
an oscillation of great frequency stability. When the L.c. circuit in the plate circuit is tuned close to
the crystal resonant frequency, steady oscillations will be established. These are maintained by C
whose oscillations value is small. By placing crystal between the gate a n ~ source, of the FET
amplifier, and feeding back a small A.c. voltage from the output, to keep crystal vibrating, the
.
circuit becomes an oscillator with precise stability. Accuracy::::: 0.01 % 'f range is 0.1 to 20 MHz.
By keeping crystal in an oven, accuracy can be improved to 0.00 I %.
C
C
·r----
G
XTAL
10M
Fig. 8.22 Crystal oscillator circuit.
8.16 FREQUENCY STABILITY
It is a measure of the ability of the circuit to maintain exactly the same frequency for which it is
designed over a long time interval. But actually in many circuits the 'f will not remain fixed but it
drifts from the designed frequency continuously. This is because of variations of number of
parameters, circuit components, transistor parameters, supply voltages temperature, stray
capacitances etc. In order to increase the stability the factors which effect the 'f largely should
be taken care of. If 'f depends only on Rand C high precision Rand C should be employed.
Also temperature compeilsating elements are to be employed.
Effect of temperature on inductors and capacitors amounts to more than 10 parts per million
per degree change.
Oscillators 449
Crystal will have mass, elasticity and damping. Crystal will have very high mass to
elastic ratio ( ~ ) and to a high ratio of mass to damping (high Q). Its value is of the order of
10,000 to 30,000.
The crystal is coupled with external C, and the LC circuit oscillates. In the oscillator circuits,
instead of external Land C, crystal is connected. i.e. crystal Land C are being used. So 'f' is fixed
by crystal itself. 'f' will not vary with temperature. To change 'f' another crystal is used.
8.17 FREQUENCY OF OSCILLATIONS FOR PARALLEL RESONANCE CIRCUIT
Total admittance
(R _ jroC)x _. 1_
JroC
Z = ----"---
(R + jroC)+ _. 1_
JroC
At resonance, imaginary part is zero
or
R + jroC
R ~ o o .
This is the expression for frequency of oscillations.
Amplitude of oscillator will be very small (v is Less) for series resonance circuit. So parallel
tuned circuits are employed.
8.18 I-MHz FET CRYSTAL OSCILLATOR CIRCUIT
In the basic configuration of oscillator circuit,
ZI+Z2 +Z3 = o.
Zl is crystal Z2 is Land C combination.
450 Electronic Devices and Circuits
Z3 is C
gd
. The frequency of oscillator essentially depends up on crystal only as shown
in Fig. 8.23. Z2 and Z3 values are insignificant compared to Zj of the crystal. Therefore the
stability is high.
.-----,...... - 22V
0-1 221lH
C
gd
ill- - - - -+-----'
C
300pF
Fig. 8.23 Crystal oscillator circuit.
When electrical input is given, mechanical vibrations are set in the crystal, due to piezoelectric
effect. But the crystal is being considered as an inductor capacitor combination only. So it acts like
a I ~ Ro L, C combination. Frequency of oscillations depend only on crystal. Other Land Cs are
insignificant, as their values are less compared to L, C and R ofthe crystal.
•
•
•
•
SUMMARY
Oscillators generate A.C. output without external A.C input by using Noise A.C
signal generated in switching. D.C power from Vee or V DD is converted to A.c.
power. The range of frequency signals generated by the circuit can be high and
output power is small.
For sustained oscillations, the conditions known as Barkhausen criterion to be
satisfied are (i) II3AI?: I (ii) Total loop phase shift must be 0° or 360° (or phase
shift of feedback network must be 180° when the amplifying device produces
another 180°).
1
In the case of JFET, R - C phase shift oscillator circuit,1o = 21tRC.j6
In the case BJT, R - C phase shift oscillator circuit, 10 = 21tRC.J6+4k where
Rc
K=-
R
• For Hartley oscillator circuit which employs two inductors and one capacitor in
1
the feedback network is,1o = l( )
21t-y L
J
+ L2 C
3
Oscillators 451
• For Colpitts oscillator circuit with two capacitors and one inductor in the feedback
• For Wein Bridge oscillator circuit, the minimum gain of amplifier must be 3. Itcan
produce variations in fo in the ratio of I 0 : I compared to a variation of3 : I in other
types of oscillator circuits.
• Thermistors with (NTCR) and sensistor with positive temperature coefficient of
resistance (PTCR) are used for frequency stability in oscillator circuits.
• The specification parameters of oscillator circuits are (i) Amplitude stability (ii)
Frequency stability (iii) Frequency range (iv) Distortion in output waveform etc.
• Crystal oscillators produce highly stable output waveform in the high frequency
range of MHz also.
OBJECTIVE TYPE QUESTIONS
1. The difference between an amplifier circuit and oscillator circuit is ______ _
2. A.C. output signal is generated by the oscillator circuits without external A.C input, by
amplifying signal.
3. Oscillator circuits employ _______ type of feedback.
4. One condition for sllstained oscillations is total loop phase shift must be
-------
degrees.
5. As per Berkauhsen criteria for sustained oscillations, II3AI must be ______ _
6. The range of frequencies over which R - C phase shift oscillator circuit is used is
7. In the Mega Hertzs frequency range, the type of oscillator circuit used IS
8. The range of frequency over which Wein Bridge oscillator used is ______ _
9. In the feedback network if two inductors and one capacitor elements are used, the
oscillator circuit is
-------
10. The oscillator circuit which employs two capacitors and one inductor in the feedback
network, is oscillator circuit.
-------
11. Naturally occuring materials used for in crystal oscillator circuits, exhibiting piezoelectric
effect are I. 2.
-------
452
12.
13.
14.
15.
Electronic Devices and Circuits
Synthetic materials which exhibit piezoelectric effect are,
1. 2. _____ _
3. 4.
The ratio of frequency variation possible with Wein Bridge Oscillator circuit is
_______ is compared to the ratio of with others oscillator
circuits.
Thermistors have temperature coefficient and the materials used are ______ _
Typical values of L, C, Rand Q of a crystal used in oscillator circuits are
L= C= R=
Q=
-------
ESSAY TYPE QUESTIONS
1. Explain the basic principle of generation of oscillations in LC tank circuits. What are
the considerations to be made in the case of practical L.C. Oscillator Circuits?
2. Deduce the Barkausen Criterion for the generation of sustained oscillations. How
are the oscillations initiated?
3. Draw the circuit and explain the princple of operaiton of R.C.phase-shift oscillator
circuit. What is the frequency range of generation of oscillations? Derive the expression
for the frequency of oscillations.
4. Derive the expression for the frequency of Hartely oscillators.
5. Derive the expression for the frequency of Colpitt Oscillators.
6. Derive the expression for the frequency of We in Bridge Oscillators.
7. Derive the expression for the frequency of Crystal Oscillators.
8. Explain how better frequency stability is obtained in crystal oscillator?
9. Draw the equivalent circuit for a crystal and explain how oscillations can be
generated in electronic circuits, using crystals.
10. Why three identical R-C sections are used in R-C phase-shift oscillator circuits?
Consider the other possible combinations and limitations.
MULTIPLE CHOICE QUESTIONS
1. Johnson Noise is due to .....
(a) Humidity (b) Atmospheric conditions
(c) Temperature (d) Interference
2. The noise that arises in electronic circuits due to variation in concentrations of
carriers in semiconductor devices is
(a) shot noise (b) thermal noise (c) Johnson noise(d) None of these
3. Expression for frequency of oscillations for the R-C phase shift oscillator circuit
using JFET is, fo =
I
(a)
2rcRC
(b) (c) (d)
2rcRCJ6
Oscillators 453
4. In the case of JFET R-C phase shift oscillator circuit, in order that AI is not less
than unity, Il of JFET must be at least
(a) 16 (b) 92 (c) 29 (d) None of these
5. ?scillator circuit in which the reactances o.
t
, 0.
2
are capacitive and 0.
3
is inductive
IS ••••
(a) Colpitts oscillator (b) Hartelyoscillator
(c) Crystal oscillator (d) None of these
6. In wein bridge oscillator circuits the range over which frequency of oscillations
can be varied is ...
(a) 3: 1 (b) 10: 1 (c) 6: 1 (d) None of these
7. Example for naturally occurring piezoelectric crystal material is ....
(a) BaTio
3
(b) Rochelle salt (c) PZT (d) None of these
8. Typical values for 90 KHz crystal are:
(a) L= 137HC=0.0235IlfR= 15 KQ=5,500
(b) L= 1 HC=0.02fR= lQQ= 10
(c)
(d) None of these
9. One characteristic feature of crystals is ...
(a) They have low mass to elastic ratio
(b) They have high mass to elastic ratio
(c) They have low Q
(d) None.ofthe above is correct
10. For Wein Bridge oscillator circuits, the minimum gain of amplifier must be
(a) 1 (b) 3 (c) 10 (d) None the these
454 Electronic Devices and Circuits
Additional Objective Type Questions (Chapter 1-8)
1. In a BJT the arrow on the emitter lead specifies the direction of ______ _
where the emitter-base junction is biased.
2. The emitter efficiency of a BJT is defined as the ratio of current of injected carriers at
_______ junction to total current.
3. type of npn junction transistor is made by drawing a single crystal
from a melt of silicon whose concentration is changed during the
crystal drawing operation by adding n or p type atoms as requires.
4. In an type of pnp transistor two small dots of indium are attached to
opposite sides of a semiconductor.
5. In the active region of common base output characteristics, the collector junction is
_______ biased and the emitter junction is biased.
6. If both emitter and collector junctions are (a) forward biased, the transistor is said to be
operating the region. (b) reverse biased, the transistor is said to be
operating in the regIOn.
7. The operation of a FET depends upon the flow of _______ carriers only. It
is therefore a device.
8. The maximurll voltage that can be applied between any two terminals of the FET is the
_______________ voltage that will cause avalanche breakdown across the
9. A MOSFET of the depletion type may also be operated in an mode.
Some times the symbol for the JFET is also is used for the MOSFET with the under-
standing that is internally connected to source.
10. The emitter diode of a UJT drives the junction of RBI and RB2 and RBB = RBI + R
B2
• The
intrinsic stand off ratio is defined as 11 = and its value usually lies,
between
-------
Oscillators 455
[ Answers toAdditional Objective Type Questions J
1. Holes or conventional current, forward
2. Emitter base, emitter
3. Grownjunction, doping
4. Alloy junction
s. Reverse, Forward
6. Saturation, Cut-off
7. Majority, Unipolar
8. Breakdown, drain-source junction
9. Enhancement, the gate
10.
RB1 + RB2
RBB
,O.S & 0.82
"This page is Intentionally Left Blank"
Appendix - I
Appendix - II
Appendix - III
Appendix - IV
Appendix - V
Appendix - VI
Appendix - VII
Appendix - VIII
Colour Codes for Electronic Components
Resistor and Capacitor Values
Capacitors
Inductors
Miscellaneous
Circuit Symbols
Unit Conversion Factors
American Wire Gaugcy Sizes and Metric Equivalents
APPENDIX-l
Colour Codes for Electronic Components
RESISTOR COLOUR CODE:
Number of zeros
(except when
Second digit , silver or g7
01d
) Tolerance
F;"t d;g;t _ I
----tm""""-=Ir"T'rT-' ,
CAPACITOR COLOUR CODE:
digit } . .
digit Capacitance In pF
ber of zeros
,...""."n,',," (%)
dc working
voltage (x 100V)
First Three Bands Fourth Band
Black - 0 Blue - 6 Gold ±5%
Brown - I Violet - 7 Silver ± 10%
Red - 2 Grey - 8 None ±20%
Orange -3 White - 9
Yellow -4 Silver 0.01
Green -5 Gold 0. 1
Colour Figure Significant Tolerance (%)
Black 0 20
Brown I I
Red 2 2
Orange 3 3
Yellow 4 4
Green 5 5
Blue 6 6
Violet 7 7
Grey 8 8
White 9 9
Silver 0.0\ JO
Gold 0.1 5
No Band 20
APPENDICES
INDUCTOR COLOUR CODE:
Inductance in IlH Decimal point or second digit ........... /
{
Decimal point or first digit -+
Number of zeros ~ ,
Tolerance (%) --_./
Color
Black
Brown
Red
Orange
Yellow
Green
Blue
Violet
Grey
White
Silver
Gold
No Band
459
Significant Tolerance (%)
Figure
0
1
2
3
4
5
6
7
8
9
10
Decimal point 5
20
COLOUR CODE MEMORY AID: W G VIBGYOR BB (W G Vibgyor BB)
Memory aid Color Number
Black Black 0
Bruins Hrown 1
Relish Red 2
Ornery Orange 3
Young Yellow 4
Greenhorns Green 5
Blue Blue 6
Violets Violet 7
Growing Grey 8
Wild White 9
Smell Silver 0.01 10%
Good Gold 0. 1 5%
460
In "'f2!1g ) « !!, ,!
\W
APPENDICES
Fig. A-i.} Relative size of carbon composition resistors with various power ratings
Specifications of Power Transistors
Device Type
PD(W) Ic(A)
VCEO{V) Vcoo(V)
hJlEMIN Max
f
T
M
.
HZs
2N6688 NPN 200 20 200 300 ~ 80 ~
2N3442 NPN 117 10 140 160 ~ 70 0,08 '
BUX39 NPN 120 30 <Xl 120 15 45 8
ECP149 PNP 30 4 40 50 30 - 2.5
Darlington Pair
2N6052 PNP 150 12 100 100 750 - 4
2N6059 NPN 150 12 100 100 750 - 4
APPENDIX-2
Resistor and Capacitor Values
Typical Standard Resistor Values <± 10% Tolerance)
n n n ill ill ill Mn Mn
10 100 10 100 10
12 120 1.2 12 120 1.2
15 150 1.5 15 150 1.5 15
18 180 1.8 18 180 1.8
22 220 2.2 22 220 2.2 22
2.7 27 270 2.7 27 270 2.7
3.3 33 330 3.3 33 330 3.3
3.9 39 390 3.9 39 390 3.9
4.7 47 470 4.7 47 470 4.7
5.6 56 560 5.6 56 560 5.6
6.8 68 680 6.8 68 680 6.8
82 820 8.2 82 820
462 APPENDICES
Typical Standard Resistor Values (± 10% Tolerance)
P' P' P' P'
~ F ~ F
~
~ F ~ F ~ F ~ F
5 50 500 5000 0.05 0.5 5 50 50 5000
51 510 5100
56 560 5600 0.056 0.56 5.6 56 5600
6000 0.06 6 6000
62 620 6200
68 680 6800 0.068 0.68 6.8
'75
750 7500 75
8000 8 80
82 820 8200 0.082 0.82 82 82
91 910 9100
10 100 1000 O.oJ 0.1 10 100 1000 10,000
110 1100
12 120 1200 0.012 0.12 1.2
130 1300
15 150 1500 O.oJ5 0.15 1.5 15 150 1500
160 1600
18 180 1800 O.oJ8 0.18 1.8 18 180
20 200 2000 0.02 02 2 20 200 2000
24 240 2400 240
250 2500 025 25 250 2500
27 270 2700 0,027 027 2.7 27 270
30 300 3000 0.03 OJ 3 30 300 3000
33 330 3300 0.033 0033 303 33 330 3300
36 360 3600
39 390 3900 0.039 0039 3.9 39
4000 0.Q4 4 400
43 430 4300
47 470 4700 0.047 0.47 4.7 47
APPENDICES
Physical constants
Charge of an electron
Mass of an electron
elm ratio of an electron
Plank's constant
Soltzman's constant
Avogadro's number
Velocity oflight
Permeability offree space
Permittivity offree space
Intrinsic concentration in silicon at 300 oK
Intrinsic resistivity in silicon at 300 oK
Mobility of electronics in silicon
Mobility of holes in silicon
Energy gap at in silicon at 300 OK
e
m
elm
h
K
1.60 x 10-
19
coulombs
9.09 x 10-
31
Kg
1.759 x lOll ClKg
6.626 x 10-
34
J-sec
1.381 x 10-
23
JfOK
K 8.62 x 10-
5
ev/oK
N A 6.023 x 10
23
moiecules/mole
c 3 x 10
8
m/sec
mo 1.257 x 10-6 Him
'0 8.85 x 10-
12
F/m
n, = 1.5 x 1010 Icm
3
r, = 230,000 W--cm
mn =1300cm2/V-sec
mp = 500 cm
2
1 V-sec
= 1.1 ev.
463
APPENDIX-3
Capacitors
Capacitance
The farad (F) is the Sf unit of capacitance.
The farad is the capacitance of a capacitor that contains a charge of 1 coulomb when the
potential difference between its terminals is 1 volt.
Leakage Current
Despite the fact that the dielectric is an insulator, small leakage currents flow between the plates of a
capacitor. The actual level of leakage current depends on the insulation resistance of the dielectric.
Plastic film capacitors, for example, may have insulation resistances higher than 100 000 MW. At the
other extreme, an electrolytic capacitor may have a microamp (or more) of leakage current, with
only 10 V applied to its terminals.
Polarization
Electrolytic capacitors normally have one terminal identified as the most positive connection. Thus,
they are said to be polarized. This usually limits their application to situations where the polarity of the
applied voltage 'viII not change. This is further discussed for electrolytic capacitors.
Capacitor Equivalent Circuit
An ideal capacitor has a dielectric that has an infinite resistance and plates that have zero resistance.
However, an ideal capacitor does not exist, as all dielectrics have some leakage current and all capacitor
plates have some resistance. The complete equivalent circuit for a capacitor [shown in Fig. A-3. 1 (a)]
consists of an ideal capacitor C in series with a resistance RD representing the resistance ofthe plates,
and in parallel with a resistance RL representing the leakage resistance of the dielectric. Usually, the
plate resistance can be completely neglected, and the equivalent circuit becomes that shown in
Fig. A-3.1(b). With capacitors that have a very high leakage resistance (e.g., mica and plastic film
capacitors), the parallel resistor is frequently omitted in the equivalent circuit, and the capacitor is
then treated as an ideal capacitor. This cannot normally be done for electrolytic capacitors, for example,
/ which have relatively low leakage resistances. The parallel Rc circuit in,
APPENDICES
c
(a) Complete equivalent circuit
---------- -------------------- - -
c
(b) Parallel equivalent circuit
Fig. A.3.1
465
(c) Series equivalent circuit
A capacitor equivalent circuit consists of the capacitance C. the leakage resistance RL in
parallel with C. and the plate resistance Ro in series with C and R
L
,
Fig, A, 3.1 (b) can be shown to have an equivalent series RC circuit, as in Fig. A. 3.1 (c). This
is treated in Section 20-6.
A variable air capacitor is made up of a set of movable plates and a set of fixed plates
separated by air.
Because a capacitor's dielectric is largely responsible for determining its most important
characteristics, capacitors are usually identified by the type of dielectric used.
Air Capacitors
A typical capacitor using air as a dielectric is illustrated in Fig. A.3.2. The capacitance is variable, as
is the case with virtually all air capacitors. There are two sets of metal plates, one set fixed and one
movable. The movable plates can be adjusted into or out of the spaces between the fixed plates by
means of the rotatable shaft. Thus, the area of the plates opposite each other is increased or decreased,
and the capacitance value if altered.
Movable
\Plates
--
Fig. A. 3.2 A variable air capacitor is made up of a set of movable plates and a set of fixed
plates separated by air
Paper Capacitors
In its simplest form, a paper capacitor consists of a layer of paper between two layers of metal foil.
The metal foil and paper are rolled up, as illustrated in Fig. A.3.3 (a); external connections are brought
out from the foil layers, and the complete assembly is dipped in wax or plastic. A variation of this is
the metalized paper construction, in which the foil is replaced by thin films of metal deposited on the
surface of the paper. One end of the capacitor sometimes has a band around it [see Fig. A.3 .3 (b)].
This does not mean that the device is polarized but simply identifies the terminal that connects to the
outside metal film, so that it can be grounded to avoid pickup of unwanted signals.
466 APPENDICES
Paper capacitors are available in values ranging from about 500 pF to 50llF, and in dc working
voltages up to about 600 V. They are among the lower-cost capacitors for a given capacitance value
but are physically larger than several other types having the same capacitance value.
Plastic Film Capacitors
The construction of plastic film capacitors is simi lar to that of paper capacitors, except that the paper
is replaced by a thin film that is typically polystyrene or Mylar. This type of dielectric gives insulation
resistances greater than 100 000 MO. Working voltages are as high as 600 V, with the capacitor
surviving 1500 V surges for a brief period. Capacitance tolerances of ± 2.5% are typical, as are
temperature coefficients of 60 to 150 ppm/oC.
Plastic film capacitors are physically smaller but more expensive than paper capacitors. They
are typically available in values ranging from 5 pF to 0.47 IlF.
Foil
(a) Construction of a paper capacitor
Band identifies
outer foil terminal
(b) Appearance of a paper capacitor
Fig. A.3.3 In a paper capacitor, two sheets of metal foil separated by a sheet of paper are
rolled up together. External connections are made to the foil sheets.
Mica Capacitors
As illustrated in Fig. A. 3.4(a), mica capacitors consist of layers of mica alternated with layers of
metal foil. Connections are made to the metal foil for capacitor leads, and the entire assembly is
dipped in plastic or encapsulated in a molded plastic jacket. Typical capacitance values range from
I pF to O.IIlF, and voltage ratings as high as 35 000 V are possible. Precise capacitance values and
wide operating temperatures are obtainable with mica capacitors. In a variation of the process, silvered
mica capacitors use films of silver deposited on the mica layers instead of metal foil.
Ceramic Capacitors
The construction of a typical ceramic capacitor is illustrated in Fig. A. 3.4(b). Films of metal are
deposited on each side of a thin ceramic disc, and copper wire terminals are connected to the metal.
The entire units is then encapsulated in a protective coating of plastic. Two different types of
ceramic are used, one of which has extremely high relative permitivity. This gives capacitors that are
. much smaller than paper or mica capacitors having the same capacitance value. One disadvantage of
. this particular ceramic dielectric, is that its leakage .r:esistance is not as high as with other types.
Another !type .ofcerarnic gives ,leakage.resistances on .the order.of 7500 MW. Becaus,e,of,its,.1ower
. ,permitiyity, this'Cetamic,produces'capacitors that are a giYen ofcap<lC}tance.
\ The range of capacitance valuesavaihiblewith·cetamic capaciforsistYilicaUy tpFto·O.I ' jiF,
, with dc working voltages up'to 1000 V. .;,
APPENDICES
(a) Construction of mica capacitor
Metal film
_ Ceramic disk
(b) Ceramic capacitor
Electrolyte
Tantalum
(c) Ceramic trimmer (d) Construction of tantalum capacitor
467
Fig. A. 3.4 Mica capacitors consist of sheets of mica interleaved with foil. A ceramic disc
silvered on each side -makes a ceramic capacitor; in a ceramic trimmer, the plates area is
screwdriver adjustable. A tantalum capacitor has a relatively large capacitance in a
small volume.
Fig. A. 3.4(c) shows a variable ceramic capacitor known as a trimmer. By means of a
screwdriver, the area of plate on each side of a dielectric can be adjusted to alter the capacitance -
value. Typical ranges of adjustment available are 1.5 pF to 3 pF and 7 pF to 45 pF.
Electrolytic Capacitors
The most important feature of electrolytic capacitors is that they can have a very large capacitance in
a physically small container. For example, a capacitance of 5000 J-lF can be obtained in a cylindrical
package approximately 5 cm long by 2 cm in diameter. In this case the dc working voltage is only
voltage is only lOY. Similarly, a 1 F capacitor is available in a 22 cm by 7.5 cm cylinder, with a
working voltage of only 3 Y. Typical values for electrolytic capacitors range from I J-lF through
100000 J-lF.
The construction of an electrolytic capacitor is similar to that of a paper capacitor
(Fig. A.3.5(a». Two sheets of aluminium foil separated by a fine gauze soaked in electrolyte are
rolled up and encased in an aluminium cylinder for protection. When assembled, a direct voltage is
applied to the capacitor terminals, and this causes a thin layer of aluminium oxide to form on the
surface of the positive plate next to the electrolyte (Fig. A.3.5(b». The aluminium oxide is the
dielectric, and the electrolyte and positive sheet of foil are the capacitor plates. The extremely thin
oxide dielectric gives the very large value of capacitance.
It is very important that electrolytic capacitors be connected with the correct polarity. When
incorrectly connected, gas forms within the electrolyte and the capacitor may explode! Such an
explosion blow§ the capacitor apart and spreads its contents around. This could have tragic
consequences for the eyes of an experimenter who happens to be closely examining the circuit when
the explosion occurs. The terminal designated as positive must be connected to the most positive of
468
gauze
(a) Rolled-up foil sheets and electrolyte-
soaked g:,lUze
APPENDICES
--.... ==== =="",/ Foil (+ plate)
======', -+-- (- plate)
Foil
(b) The dielectric is a thin layer of
aluminium oxide
Fig. A. 3.5 An electrolyte capacitor is constructed of rolled-up foil sheets separated by electrolyte-
soaked gauze, the dielectric is a layer of aluminium oxide at the positive plate
the two points in the circuit where the capacitor is to be installed. Fig. A. 3.6 illustrates some circuit
situations where the capacitor must be correctly connected. Nonpolarized electrolytic capacitors can
be obtained. They consist essentially of two capacitors in one package connected back to back, so
one of the oxide films is always correctly biased.
Electrolytic capacitors are available with dc working voltages greater than 400 Y, but in this
case capacitance values do not exceed 100 mF. In addition to their low working voltage and polarized
operation. Another disadvantage of electrolytic capacitors is their relatively high leakage current.

ac
voltage
(c) Connected between + 5.7 V and a grounded ac voltage source
Fig. A. 3.6 It is very important that polarized capacitors be correctly connected. The capacitor
positive terminal voltage must be more positive than the voltage at the negative terminal.
APPENDICES 469
Tantalum Capacitors
This is another type of electrolytic capacitor. Powdered tantalum is sintered (or baked), typically into
a cylindrical shape. The resulting solid is quite porous. so that when immersed in a container of
electrolyte. the electrolyte is absorbed into the tantalum. The tantalum then has a large surface area
in contact with the electrolyte (Fig. A. 3.5). When a dcforming voltage is applied, a thin oxide film
is formed throughout the electrolyte-tantalum contact area. The result, again, is a large capacitance
value in a small volume.
Capacitor Color Codes
Physically large capacitors usually have their capacitance value, tolerance and dc working voltage
printed on the side of the case. Small capacitors (like small resistors) use a code of colored bands
(or sometimes colored dots) to indicate the component parameters.
There are several capacitor color codes in current use. Here is one of the most common.
Color Significant Figure Tolerance (%)
Black 0 20
Brown I I
Red 2 2
Orange 3 3
Yellow 4 4
Green 5 5
Blue 6 6
Violet 7 7
Grey 8 8
White 9 9
Silver 0.01 10
Gold 0.1 5
No band 20
A typical tantalum capacitor in a cylindrical shape 2 cm by I cm might have a capacitance of
100 mF and a dc working voltage of20 V. Other types are available with a working voltage up to 630
V, but with capacitance values on the order of 3.5 mF. Like aluminium-foil electrolytic capacitors,
tantalum capacitors must be connected with the correct polarity size of the inductor, the maximum
current can be anything from about 50 mA to I A. The core in such an inductor may be made
adjustable so that it can be screwed into or partially out of the coil. Thus, the coil inductance is
variable. Note the graphic symbol for an inductor with an adjustable core [Fig. A. 3.6(b)].
APPENDIX-4
Inductors
Magnetic Flux and Flux Density
The weber* (Wb) is the Sf unit of magnetic flux.
The weber is defined as the magnetic flux which. linking a single-turn coil, produces an emf of
1 V when the flux is reduced to zero at a constant rate in 1 s.
The, tesla* * * (T) is the Sf unit of magnetic flux density.
The tesla is the flux density in a magnetic field when 1 Wb of flux occurs in a plane of 1 m
2
;
that is, the tesla can be described as 1 Wblm
2
.
Inductance
The Sf unit of inductance is the henry (H).
The inductance of a circuit is 1 henry (1 H) when an emf of 1 V is induced by the current
changing at the rate of 1 Als.
Molded Inductors
A small molded inductor is shown in Fig. A. 4.2(c). Typical available values for this type range from
1.2 J,lH to 10 mH, maximum currents of about 70 rnA. The values of molded inductors are identified
by a color code, similar to molded resistors. Fig. A. 4.2(d) shows a tiny-film inductor used in certain
types of electronic circuits. In this case the inductor is simply a thin metal film deposited in the form
of a spiral on ceramic base.
Laboratory Inductors
Laboratory-type variable inductors can be constructed in decade box format, in 'Which precision
inductors are switched into or out of a circuit by means of rotary switches. Alternatively, two coupled
coils can be employed as a variable inductor. The coils may be connected in series or in parallel, and
the total inductance is controlled by adjusting the position of one coil relative to the other.
APPENDICES
Color Code For Small Inductors

Coil ,..... . __ '
) Ferrite pot core
B bb
' .,K /
o In I
471
Fig. A. 4.1 Some low-current, high-frequency inductors are wound on bobbins contained in a fer·
rite pot core. The ferrite core increases the winding inductance and screens the inductor
Coil to protect adjacent components against flux leakage and to protect the coil from external
magnetic fields. The coil is wound on a bobbin, so its number of turns is easily modified.
Three different types of low-current inductl)rs are illustrated in Fig. A. 4.2. Fig. A 4.2(a)
shows a type that is available either as an air-cored inductor or with a ferromagnetic core. With an air
core, the inductance values up to about 10 mH can be obtained. Depending on the thickness of wire
used and the physical.
472
(a) Inductor with air core or
ferromagnetic core
(c) Molded inductor
/'t..
1 1
II
1 1
II
1 1
II
APPENDICES
(b) Circuit symbol for an inductor with
an adjustable ferromagnetic core
(d) Thin-film inductor
Fig. A. 4.2 Small inductors may be wound on an insulating tube with an adjustable ferrite core,
molded like small resistors, or deposited as a conducting film on an insulating material
If the mutual inductance between two adjacent coils is not known, it can be determined by
measuring the total inductance of the coils in series-aiding and series-opposing connections. Then,
and
Subtracting,
Therefore,
L=L+L+2M
a I 2
L=L+L - 2M
b I 2
L - L = 4M
a b
M = k ~ L I L 2
for series-aiding
for series-opposing
From these two equations, the coefficient of coupling of the two coils can be determined.
Stray Inductance
Inductance is (change in flux linkages) / (change in current). So every current-carrying conductor
has some self-inductance, and every pair of conductors has inductance. These stray inductance are
usually unwanted, although they are sometimes used as components in a circuit design. In dc
applications, stray inductance is normally unimportant, but in radio frequency ac circuits it can be
considerable nuisance. Stray inductance is normally minimized by keeping connecting wires as short
as possible.
APPENDICES 473
-------------- Summary of Formulae --------------
Induced emf
Induced emf
Inductance
Inductance
Flux change
Self - inductance
Mutual inductance
Induced emf
Mutual inductance
Mutual inductance
Mutual inductance
Energy stored
Energy stored
Inductances in series
Inductances in parallel
Total inductance (series-aiding)
Total inductance (series-opposing)
Mutual inductance
.1<1>
e =--
L ll!

e = --
L ll!

MIll!

L = --

A
= J..l, J..l
o
N t
? A
L = J..l J..l N- -
, 0 t

.1i I

e =--
L ill
M =
M
M = k
M
M=
I
W = - U
2
2
B2AI
W=--
2J..lO
L = L+L + L + .....
5 I 2 3
I I I I
- =-+-+-+ .....
Lp LI L2 L)
L = L +L +2M
I :2
L = L +L - 2M
I 2
Lo -Lb
M = ---"'-
4
APPENDIX-5
Miscellaneous
Ionic Bonding
In some insulating materials, notably rubber and plastics, the bending process is also covalent. The
valence in these bonds are very strongly attached to their atoms, so the possibility of
current flow is virtually zero. In other types of insulating materials, some atoms have parted with
outer-shell electrons, but these have been accepted into the orbit of other atoms. Thus, the atoms are
ionized; those which gave up electrons have become positive ions, and those which accepted the
electrons become negative ions. Thi s creates an electrostatic bonding force between the atoms,
termed ionic bonding. Ionic bonding is found in such materials as glass and porcelain. Because
there are virtually no free electrons, no current can flow, and the material is an insulator.
Insulators
Fig. A. 5.1 shows some typical arrangements of conductors and insulators. Electric cable usually
consists of conducting copper wire surrounded by an insulating sheath of rubber or plastic. Sometimes
there is more than one conductor, and these are, of course, individually insulated.
Conductor '''--''',1
Insulation --..... _. II
Ii
:'
Protecti ve _
----
sheath
Printed copper
conductor
Insulation
Fig. A. 5.1 Conductors employed for industrial and domestic purposes normally have stranded
copper wires with rubber or plastic insulation. In electronics equipment, flat cables of fine wires
and thin printed circuit conductors are widely used.
Conductors
The function of a conductor is to conduct current form one point to another in an electric circuit. As
discussed, electric cables usually consist of copper conductors sheathed with rubber or plastic
APPENDICES

/A/ ,/'
Screened
coaxial
cable Conductor Screen
Circular
multi conductor
cable
Flat
multiconductor-
cable
475
Fig. A. 5.2 Many different types of cables are used with electronics equipment.
insulating material. Cables that have to carry large currents must have relatively thick conductors.
Where very small currents are involved, the conductor may be a thin strip of copper or an
aluminium film. Between these two extremes, a wide range of conductors exist for various applications.
Three different types of cables used in electronics equipment are illustrated in Fig. A. 5.2 conductor
and a circular plated conducting screen, as wen as an outer insulating sheath. The other two are
multi conductor cables, one circular, and one flat.
Because each conductor has a finite resistance, a current passing through it causes a voltage
drop from one end of the conductor to the other (Fig. A. 5.3). When conductors arel long and/or
carry large currents, the conductor voltage drop may cause unsatisfactory performance of the equipment
supplied. Power (12 R) is also dissipated in every current-carrying conductor, and this is, of course,
wasted power.

I
(a) Current flow through a conductor produces a voltage drop along the conductor

All rights reserved. No part of this book or parts thereof may be reproduced, stored in a retrieval system or transmitted in any language or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publishers.

Before electronic engineering came into existence.A.Thomson proved the same experimentally in 1897.J. J. we study the applications of these laws to human needs. Electrical engineering mainly deals with motion of electrons in metals only. Two years later. Braun built the first tube. a semiconductor diode was fabricated but they could not succeed. in electrical engineering the voltages and currents are of very high-kilovolts. whereas Electronic engineering deals with motion of charged particles (electrons and holes) in metals. it is KHzs. whereas in electronic engineering one deals with few volts and rnA.
. electrical engineering flourished. semiconductors and also in vacuum. and Amperes.. based on the motion of electrons. Electronics is the science and technology of the passage of charged particles in a gas or vacuum or semiconductor.was made in 1895. Yet another difference is.Brief History of Electronics
In science we study about the laws of nature and its verification and in technology. So. in electrical engineering. In 1906. Lorentz postulated the existence of discrete charges called electrons. in making it work. In the same year. and called it
Cathode ray tube (CRT). the frequencies of operation are 50 Hertzs/60 Hertzs.
In 1904. Another difference is. when H. (high frequency). whereas in electronics. GHzs. Fleming invented the Vacuum diode called 'valve'. semiconductor technology met with premature death and vacuum tubes flourished. The beginning for Electronics. MHz.

with more than 1000 but < 10. So in 1945. it generated lot of heat. 1947: Shockley . in 1884 Institute of Electrical Engineers was formed and in 1963 both institutes merged into one association called IEEE (Institute of Electrical and Electronic Engineers).1000 components/chip was developed. In 1950. The filaments get heated to > 2000° k. 1959: Integrated circuit concept was announced by Kilby at an IRE convention.000 components per chip.
Major milestones in development of Electronics:
1895: H. The first radio broadcasting station was built in 1920 in USA. C components Radio. 1969: LSI. IC .Postulated existance of Electrons 1897: J. chip with 1000 Transistors (4004m) 1971: 4 bit Microprocessor was made by INTEL group. FFs were developed.
. 1975: VLSI: Very large scale integration> 10.Complimentary High Metal Oxide Semiconductor ICs were announced by INTEL group.Proved the same 1904: Fleming invented Vacuum Diode 1906: De Forest developed Triode 1920: Radio Broadcasting in USA 1930: Black and White Television Transmission in USA. device was announced. De Forest put a third electrode into Fleming's diode and he called it Triode. Before that. (BJT) 1950: Colour Television Transmission started in USA. black and white television transmission started in USA. 1970: INTEL group announced. small change in grid voltage produces large change in plate voltage in this device.000 components per chip (integrated or joined together). Thomson .In 1906 itself. control systems Computers
Vacuum Tubes ruled the electronic field till the invention of transistors. The filaments get burnt and tubes occupy large space. LOGIC GATES. Television.invented the junction transistor. 1975: MSI (Multiplenum. A. Solid State Physics group was formed to invent semiconductor devices in Bell Labs.J. L. R.Large Scale Integration. In 1930. ICs. Address) 100 . landline communications Industrial electronics. ICs were made. 1969: SSI 10 . The difficulty with vacuum tubes was.100 components/chip. Colour television broadcasting was started. Telephone . USA. so that electron emission takes place.
A
In 1912 Institute of Radio Engineering (IRE) was set up in USA to take care of the technical interests of electronic engineers. The electronics Industry can be divided into 4 categories: Components Communications Control Computation Transistors. 1975: CHMOS . Lorentz .wireless.

The Mathematical Equations describing the Motion are derived. Magnetic Fields and combined Electric and Magnetic Fields are given..
• •
. The Practical Application of this study in a Cathode Ray Oscilloscope is also given.namlCS
•
In this Chapter.
•
The path or trajectories of electrons under the influence of Electric Fields.

considering electron charge as e.. First we consider only uniform electric fields and then uniform magnetic fields. For large scale phenomena. The charge of an electron is 1. we study the motion of electrons in electric fields and magnetic fields.6 x 10.11 x 10-31 Kgs..1
There are two different types of Electron Models. The radius of an electron is estimated as 10-15 metres and that of an atom as 10-10 metre. such as. force = 1 x E Newtons. ( 1.
. Classical Model 2. In this chapter..1tes that force is opposite to the direction of E..
.2
Electronic Devices and Circuits
ELECTRON DYNAMICS The term Electron Dynamics refers to the analogy between an electron under electric and magnetic
fields.
1.
1. and a body falling under gravity... The assumption that electron is a tiny particle possessing definite mass and charge.. Wave-Mechanical Model satisfactorly explains those phenomena. electron behavior in a crystal or in an atom.. is the Classical Model. For a positive charge 'q'. the trajectory of the electron in the electric field can be found out. it traverses a path and falls under gravity..1 THE FORCE ON CHARGED PARTICLES IN AN ELECTRIC FIELD
The force experienced by a unit positive charge at any point in an electric field is the electric field intensity 'E' at trat point.2 MOTION OF CHARGED PARTICLES IN ELETRIC AND MAGNETIC FIELDS 1.2. F = q x E Newtons where F is in Newton's. classical theory results do not agree with experimental results. Classical Model satisfactorily explains the behavior of electrons in electric and magnetic fields.1 )
For accelerating potential. Its units are V1m For unit pusitive charge. the force. :. Wave-Mechanical Model.19 Coulombs. If a shell is fired from a cannon. 1.
a=dt
dv
By solving this equation. while the assumption that electrons travel in the form of waves is called the Wave-Mechanical Model. and E in V1m But by Newton's Second Law of Motion. in the subatomic systems. We shall now consider the trajectories of electrons under different conditions.
F=-eXE
In this case negative signindic. These are very small and hence all charges are considered as Points of Mass. such as. The motion of an electron is similar to the trajectory of a shell. q is in coulombs. But. F = m x a and F = q x E
mx dt
dv
= qXE
. The mass of an Electron is
9. parallel electric and magnetic fields and then perpendicular electric and magnetic fields. electron transaction in a vacuum tube Classical Model gives satisfactory results..

. Let 'd' be the distance between the plates.. acceleration in that direction is zero.A is at a positive potential + Va with respect to B.
Ex
=
-v
(X-Xo) -
-v
x
Negative sign is for work done on a positive charge.. The acceleration along x direction is also zero. x = fvox.. (1. The integral gives the work done. z x=O} y= 0 at t z=0
v--------zt-.dt
dt
dx
= vox
x=voxxt . there is no force in the z direction. incremental change is to be considered.2. If V is not uniform.. V = 0. vy=ayxt dy .. Let an electron enter the plates at point 0.3 Two Dimensional motion.__
d
=0
Fig 1.. against the field. .5 Two DIMENSIONAL MOTION
LetA and B be two parallel plates.
E=
dv
dx
1. 1. For electrons the electric field E = + V/d Butthis is true when V and 'd' are small and V is uniform. .8
Electronic Devices and Circuits
Negative sign is to indicate that the work is done against the field..3 ) So what is the motion of the particle? v x = velocity in the x direction The initial conditions Vx = vox' Vy = 0.direction.A
~. 0
____.: There is no electric field along the x-direction..and
y m
=
V
d
.7) As the field is acting downwards. So velocity along the x direction is constant vox = V x .= v = ay t x dt y Y= But
f
a y tdt
= ay x 2
E
y
t2
eEy a = . so the component of velocity in that direction remains constant. there is constant acceleration along y . with initial velocity vox ( Fig.. + 8 .
Since..

.9 )
I
The trajectory ofthe electron is as shown in Fig.Electron Dynamics and CRO
13
~=
2Vo .
If an electron takes a time T sec to travel a distance of L m in the conductor.6 For problem 1. energy is IOev..l.Q
B
A
1 T
e = 45
0
Fig 1. directed as shown. ( 1. Solution
~=
2V0 V x d x Sin 28 Since.4. the total number of electrons passing through any cross section of wire in unit time is : Total Number of electrons in a conductor = N
. which will cause the electron to reach point B.
----------'-.p
-4-5-0-r--:l"'---2-c-m__
~
so
IO~v
--rI--r..5..
Ym is the maximum displacement in the 'y' direction and ~ is the maximum displacement in 'x' direction.
V= ?
V
~
= 2 cm
d=?
2= 2xlO xdxxSin900
V = 10d Volts
1. J C..6 has an initial velocity due to an energy of 10 eV. 1. P and Q are conducting plates..
Assuming a uniform current distribution.2.
Problem 1..4
The electron shown in Fig. V.
Ne
where N is the total number of electrons contained in a conductor of length L.S. d Sm20
. Find the potential V to be placed on electrodes P and Q. But 1=
=
A ' where J is amp/m 2 and A is the
I
T.
V0 = 10 volts. It is defined as the current per unit area of the conducting medium..A (m 2)..6
CURRENT DENSITY
It is denoted by J.

..10)
Ne
I J=
But
~~
I
Distance L .... N Thus. I =
T
T'
.. n= LA IJ=nevl But n x e = charge density in Coulombs/m3 = p p the charge density is the electric charge in coulombs per unit volume ( m3 ) v is velocity in m/sec J:--=-p-v----.14
Electronic Devices and Circuits
Ne
The charge per second passing through any point = But rate of change of charge is current.e...
But L
x
Nev J= AL A is the volume of the conductor. containing N electrons.
::~
N
. =VelocIty v
v
J is the Current Density in Amp / m2 Time =
L T can be replaced by -..
'-:1
I
J= A
I=T
Ne
Ne
A = Cross sectional Area of the conductor
J= AxT
L T= v
/J= / A x L = volume ofthe conductor
A x L = Electron Concentration per unit volume 'n' . electrons/m3. (1.
L:
gives the electron concentration = n
Nim3
Total Number of electrons in a conductor = N i..
SO..

11)
1. Electron velocity making an angle '9' with the field: Motion is helix..... assume that I and B are perpendicular to each other.·
. I in amps. F=BIL= But BxNe xL T
Ne
L T = v velocity in m/sec.2.2.
Case (ii) Case (iii) Case (iv)
Electron-moving parallel to the field: No effect... of free electrons in a conductor L = length of the conductor in meters T = The time taken by the electron to travel a distance ofL meters Total no. carrying current I is situated in a magnetic field of intensity B wb/m 2 or tesla. Fm = Bev Newtons
1.... B and v should be perpendicular to each other.8. Electron moving perpendicular to the field: Motion is a circle..12 )
I
.Electron Dynamics and CRO
J=nxexv=pxv n x e = Charge Density p in Coulombs I v = Velocity in m/sec..8
MOTION IN A MAGNETIC FIELD
Case (i)
Electron at rest: No effect .... only when B and v are perpendicular to each other. B is the magnetic field strength in Webers per unit area ( m2 ) In the above equation 1... the force F acting on the conductor is IF m = B x I x L Newtons 1. J = Current Density in Amp/m2. F = Bev v = 0
m '
Fm=O ~ will be there. Let N = No.7 FORCE IN A MAGNETIC FIELD If a conductor of length L. m3
15 . of electron passing through conductor in unit time = NIT Rate of flow of charge = Ne
T
This by definition is current I.
Force experienced by each electron due to magnetic field...
I=T·
Force due to magnetic field is. (1. (
Where B is in wb/m2. L in meters and F is in Newtons..

. v and B are constant in magnitude since fm is constant in magnitude and perpendicular to the direction of motion of the particle.in magnetic field. force acting on the particle or electron in the y-axis.
y
~
I
I
B'
Fig 1. a particle moving in a circular path with a constant speed v. m. velocity of the electron along x .axis) perpendicular to paper.axis.=eBv R R= mv Be v = J2e v o m R= =F
~J2evo
B. 7 Motion. This type offorce resolution is motion in a circuluar path with constant speed. To find the radius ofthe circle. twisted and related the motion of the mass.e m
=
.37 X lO-6 RB If
v Vo
rv .
B
v
f
flux density wb/m2 .
Distance --Velocity
T = period of rotation.J2m Vo B e
_ 3. direction being along the neutral axis (z ..!.16
Electronic Devices and Circuits
A particle whose initial velocity has no component perpendicular to a uniform magnetic field will continue to move with constant speed along the lines of the flux. Let. Then.
The path described is a circle since it is analogous to a mass tied to a rope. mv 2 . has an acceleration toward the center of the circle of magnitude v 2/R where R is the radius ofthe path in meters.
R T=21tv
Circumference Velocity
.

Determine the distance between this initial path axis and the point at which the electron leaves the field region assuming that all the trajectory is within a vacuum. the motion is helix because of v Cos e electron moves in that direction in a straight line.In
e -_ 3. ~[~2~VO 1 SinS
R
=
-1 B
~m e
" -V'Va S.
.. The field region is confirmed betwen two parallel planes. T The distance covered along the B direction in one revolution is called the pitch of the helix.37 x 10B
16
l-v vain e
rrv]s·
T = Period of rotation 21tR T=-vSine mvSine R=--Be 21tm vSin e 21tm T=----=-Bev Sine Be T
=
B
35.
T= R=
21tR vSin e m v Sin e eB
21tm T=eB
P=
~2e Vo = .sec.5
An electron initially at rest is accelerated through a 2 KV and then enters into a region in which a magnetic field of flux density 0.2 x 10-6 v va· C os e B Therefore. v Cos e 21tm -m
eB
21.5
picoseconds (p.18
Electronic Devices and Circuits
R~ : . 3 cm apart perpendicularly to the initial path.03 wb / m2 is maintained. p = v Cos e.. it will describe a circle. _ -
rv
Problem 1. and because ofv Sin e.)
IfP is the pitch of the helix.

Electron Dynamics and CRO
19
Solution E is the point where the electron enters the magnetic field.9 For Problem 1..:. If the distance between A and B is 5 cm.
R=5 cm .11 For Problem 1.= B ev
R
mv
2
or
mv R=-
.
OA=
B
o
1.6.03
Fig 1. OAC is a right angled triangle... A voltage pulse as shown in Fig..01 f. EA is the arc of the circle of radius OA.1 O.
.---. OC = 3 cm AC =4 em CD = OE = Radius of the circle OA But CD = 5 cm AD = 5 cm .6xlO19
I x-0. calculate the transit time of the electron ... and the plates could be assumed to have a geometry starting from the fundamentals.
eB
v=
R=
~2:V
~~2ev =~2mv .l see
Fig 1..
A
100 V
'so
B
oCIi<. Problem 1.lxI0-31 x2x103
v
eB Radius of the circle.
Fig 1.6.~
t = 0
0. OA = 5 em.5 em
----...4 cm = I cm The centre should lie along the OE only since E is a part on the circle and A is also a point of the circle.
. .
. EA is the path of electorn because of the effect of magnetic field.
.Therefore the centre cannot lie any where else except along OE.5. 1. OAC is right angled triangle OC = 3 cm.~
m e
2x9..--..10 For Problem 1.11 of amplitude 100 volts and direction of 0..I. _ _~~ .01 x 10-6 sec is applied so that electrode A becomes positive with reference to B. __ .----s"'witeh-----il
/11---.. AD ? v
E .6 An electron finds itself at rest at electrode B as shown in Fig.

..13 ). when Cos rot = -1
2a
Ymax = 002
at
~ax= ~
...
.2. vy=voy-at Where 'a' is the acceleration of the electron and 'a' = E. then also the effect of magnetic fieldis nil (Since. m ( Negative sign is due to the direction of acceleration which is opposite to that of electric field ). So resultant motion is due to e only. 1. Then the equation of motion of the electron is. x is max.11 PARALLEL ELECTRIC AND MAGNETIC FIELDS If e and B are parallel to each other and initial velocity ofthe electron
is zero.19)
Ix=.Cos rot)
a x = . The resulting motion solely depends upon 'e' ( Fig. (cot-Sin cot) I
Hence the motion is cycloidal. the magnetic field exherts no force on the electron. dy
dt
=
Voy . If the electron is possessing initial velocity v0 and is along the magnetic field. 13 Parallel electric and magnetic fields. voy is parallel to 8). when rot = 21t
~ax
=
a
00 2
(21t ..(1 ..Cos rot)
00
Integrating to get x.0)
=
21ta
00 2
1.. y is max. x=
~(t-~)
.Electron Dynamics and eRO
23
But
x = roy
x = ro
[:2 ]
(1 .e
&
y
f
o
~B
x
Fig 1... (1...at
.
x is max when Sin rot = 0
x is max where Sin rot is 0 and ro is max.

A
•c
Fig 1. i. Determine the minimum distance from the source at which an electron with zero velocity will again have zero velocity.l:
. y
=
v oy t . the distance travelled along y-axis increases with each revolution._ . e. x
=
-. So the resulting motion is helical with a pitch that changes with time. And.· 24
Electronic Devices and Circuits
Integrating. will have again zero velocity at B (FigI. The expression for x.3
SIMPLE PROBLEMS INVOLVING ELECTRIC AND MAGNETIC FIELDS ONLY
Problem 1. The 'Radius' of the circle is independent of E. Therefore.l:
m
a=If initially a component of velocity vox is perpendicular to B.Sin wt]
00
AB = Be = Xmax
a The expression for Xmax = 2 x 27t
ro
27t .14 For Problem 1. . From Eg.=t
ro
a=-=m. The magnetic flux density is 0.:.10 ).
Solution
When the electron is under the influence of perpendicular magnetic and electric fields..9.-[oot .d m
eV
e.0 I wb/m 2 and the electric field strength is 104 V1m. the velocity along the field changes with time.e. An electron emitted with zero velocity at A. its motion is cycloid. AB is the minimum distance from the source of electrons (A) where the electrons will have again zero velocity. because of the effect of electric field E.9
A point source of electrons is situated in mutually perpendicular uniform magnetic and electric fields.14).Y2 at2 .2. ( 1. the electron will describe r'rcular motion. the distance travelled by the electron is ( See Section 1.19)..
1.

4 PRINCIPLES OF CRT
Electronic Devices and Circuits
1. These are basically very fast x .
Vertical amplifier Horizontal amplifier Time base Circuit Power supplies
5. The voltage under examination is applied to the y-axis or as vertical input of the eRO. eRO is an extremely useful and versatile laboratory instrument used for the measurement and analysis of waveforms and other phenomena in electronic circuits. Double Beam/ Dual Beam 3. Storage oscilloscope. 4.16.2 TYPES CRO's
OF
1. the display appears as a stationary pattern on the eRG.1 BASIC CRO CIRCUITRY: A eRO consists of
1. 2. 1.4. Dual trace 4. Block schematic of Cathode Ray Oscilloscope (CRO). Cathode Ray Tube The block schematic is shown in Fig.
Ext. 3.
1.axis or horizontal input is an internally generated linear ramp voltage or time base signal. Horizontal Output
II
~--------~11l
Fig 1.
. Single Beam 2.16.y plotters.26 1. In the usual eRO application.4. When the input voltage is repetitive at fast rate. the x .

solid state light emitting arrays ofGaAs diode and plasma cells. velocity. to enable electrons to reach screen.
1. The cathode potential is several thousand volts negative.
. there is a small hole or aperture.. where as astigmatism control is used to make the dot as round as possible. CRT is the heart ofCRO.
This is analogous to a Vacuum Device Triode: a cathode supplying electrons. The Intensity of the beam on the screen can be easily controlled by varying the grid voltage. The operation of CRT may be described by the five regions mentioned below ( Fig. Beam Deflection Region The beam is positioned to the desired 4. 5. electrons are liberated with considerable energy.17 Cathode
1.ayer of Barium and Strantium coated on it
Anode
Accelerating Anodes (2000 v -10 Kv )
Vertical Plate Ilorizontal Plate
~ay
Fig 1.4.17 ) : 1. in both the grid and anode. x-y co-ordinates position Beam Target Region Display Screen.
Graticules
J--:--
Cathode Nickel Cylinder with a l.4
BEAM
Focus
It contains the beam focusing electrodes which change the beam pattern as a small round dot. The two electrodes are focus and astigmatism electrodes. a grid controlling their rate of emission. with respect to anode so that. Focus control electrode adjusts the beam to be concentrated as a dot.
Grid
Focusing Anode Final Apperture
1 """". c--------------------------------------/
T
.. and an anode collecting them.4. In a CRT. 1. Beam Focus Section The beam is focused Space potential controls the beam Beam Post Acceleration Region 3. Beam Generating Area The electron beam is generated by Thermionic Emission 2.Electron Dynamics and CRO
27
CRO nowadays is being replaced by electro luminescent panels.3
BEAM GENERATION
Tube (CRT). permitting a narrow beam to emerge from the anode.

8 GR\TICliLES
These are the scale markings on the CRT Screen.4.
1. to obtain high electron emission at moderate temperatures.5 screen . than does electrostatic deflection. i. Internal Graticule 3. They are of three kinds 1. At high frequencies inductors with few turns are necessary to obtain fast current changes.Calcium Tungstate gives blue colour.
1. Intensity knob controls the negative voltage of the grid.OOOY. Projected Graticule is provided with some covers and allow greater flexibility in graticule pattern.
. The grid is biased negative with respect to cathode. It controls the density of electrons being emitted from the cathode.4. Inductive reactance increases with frequency. Electromagnetic Deflection System has a substantial cost advantage.5 BEAM DEFLECTIO"l
Electronic Devices and Circuits
Many CRTs differ only in this method. Magnetic deflection allows a wider beam deflection angle. This is done by changing current levels in an inductor. Magnetic Deflection is accomplished by changing magnetic field. Thus TV sets and medical monitor use Electromagnetic Deflection.4. When the beam hits the phosphor.28
1. the voltage between cathode and plate is approximately 4 KY.e. In electrostatic CRTs. The original accelerating field is the potential between cathode and deflection plates. It has the capability of converting electrical energy into light energy.7 CRT DISPLAY SCREEN
Phosphor is the usual read out material on the target. The phosphor is classified as (1) Short persistence (decay in less than 1m sec) (2) Medium persistence (2 sec) (3) Long persistence (minutes).RATION
This is important for writing speed. External Graticule 2.
1. The accelerating field bends the beam towards the axis and thus changes the waveform display or decreases deflection sensitivity. a fluoresence or light emission is observed. yellow to green region.4. Voltages applied to the acceleration anode vary from 250V to 10. Projected Graticule External Graticule is screened outside the CRT screen. Fluorescent screen material is Zinc Orthosilicate ( P-1 phosphor) P . Cathode in CRTs is a nickel cylinder with a layer of barium and strontium oxide deposited over it.6 BEAM POST ACCEL~.. The Internal Graticule is screened inside the CRT screen. Two phenomena occur when a phosphor is bombarded with a high energy electron beam. For this a resistance spiral is used inside the tube envelope on which is impressed an accelerating voltage of 10 KY. human eye tends to peak in ~ 55000 AO. Iffull screen beam deflection bandwidth desired is less than 20 KHz. When the excitation beam is removed phosphorescence remains for sometime and indicates where the phosphor had been stimulated into light emission.

The switch can be a BJT. 1.
v
~t
Fig 1. Thus a P. Many AC signals are functions of time.4.C Network. a ~ing. Acqudag coating is a Graphite coating.lI1ode is at 1200V.18.5 1. 1. CI hi~ switch alternati\ el)' connects each vertical channel of CRT. Switching can be done at a faster rate when electronic devices are used as switches. Time bases generate an output voltage which is used to move the beam across CRT screen in a straight horizontallme. and accelerating anode is at 2000V. A typical case is. The end result is that all electrons entering the lens area tend to come together at a point called the focal point. Dual trace CRO enable the portrayal of two vertical beams. and return the beam quickly back to its starting point. It is at a po.>itive potential and returns secondary electrons to the cathode. ith high velocity and there will be secondary emission. It returns rapidly from right to left and so cannot be perceived. In CRO focussing is done by varying the focusing electric voltage since the electrons arrive at the screen v.1
DEFLECTION SENSITIVITY
Fig.
1.
ELECTROSTATIC DEFLECTION SENSITIVITY
Electrostatic deflection sensitivity of a pair of deflection plates of a Cathode Ray Oscilloscope (CRO) is defined as the amount of deflection of electron spot produced when a voltage of I V DC is applied between the corresponding plates. after each sweep.
1.9 TI!\IE BASE
This permits an operator to display voltage or current variations in time. C discharges.\' (lilt! eRO
29
The rocu~ing anode and accelerating anode form electrostatic lens.
. An electron passing through these anodes has two forces acting on it.18 ( b) Time base wave form. When S is closed.and so appears as a line. with an electronic switch. These electrons are attracted 0) the acqudag coating and returned to the cathode. focusing . The high accelerating voltage attracts the electrons and speeds it up in a forward direction.D of 800V will produce strong cb:trostatic field. The left to right or forward movement of the beam is called trace interval or forward trace. So it is an essential feature. The sweep action is achieved by a saw tooth wave form. When the switch S is open C gets charged. It consists of a single beam ('RT. JFET or any other electronic switching device.5. From left to right. to represent time varying functions. and the electrostatic lield between the t\\'o electrodes tends to deflect the electron. producing a saw tooth wave form. the beam moves slov.Electrol1 DYIl(lmic.18 (a) R .le time base generator and t\\O identical vertical amplifiers.l).
Time Base Circuit
A typical circuit to deflect electron beam along x-direction on the screen is as shown in Fig.

Even after decreasing s.Electron Dynamics and CRO
31
The electron follows a parabolic path from A the point of entrance. the same problem will be there. to get optimum value of deflection sensitivity..:sx m X y2 If e is the angle with the axis that the electron beam makes after emerging out from the field of the deflection plates. Let Then But
D = D2 +
2"
1
Vd xexl xD d = ---"------.: upward velocity at t = 0
Vd x e x 12 YS x --"----.. Therefore deflection 'd' of the spot can be changed by
changing the ratio Vd ..... but then the electron may hit the plates.s x me x y2
y2
= 2xex Va
m
V 10 d=-fl x Va 2s for a given CRT. but the size of CRT becomes large. Then d I = ut + YSat2
d I = Y2
=
X
ax (
~
r
d2
O2
ut = 0
1 t= Y
.. /. Va SE = Deflection Sensitivity =
v.... the point of leaving the field.
. and s are fixed. during this period be d I .. So the only alternative left is by reducing Va' So a compromise has to be made in the design of CRT.. ( 1..= ~
d
10
cmN
. Let the vertical displacement.20 )
IxD
Deflection Sensitivity SE ( with Electrostatic Field) can be increased by increasing /. Then
Tan
e=
-
yy
y
= -
or d = D x 2 2
y
yy
d = Vd x ex 1x O2 2 s x m xv 2
Total deflection. to point B. We can increase D. D.

. If we assume a small angle of deflection OM::::: I
8= -
I
R
. Let.-0 . where R is the radius of the circle.
v =
o
~2evo
m
I = length of the magnetic field L = Distance between circle of the field and the screen y = Total y deflection B = flux density in wb/m 2 Vo = initial velocity V 0 = accelerating voltage SM = Deflection sensitivity due to magnetic field = m/wblm2 = m 3 /wb The path OM will be the arc of a circle whose centre is at Q.....(
1
L
:~/~I
Fig 1.
Cathode
IA I
°·
• ·1· . If the magnetic field points out of the paper. the electron experience a force ofmagnitude e B Vo where Vo is velocity. OM = R8..... .. .. the beam is deflected upward.
p'
Q
•
p'
Jr'------c
. f-Z--p
8
:0' 1
1 1
• 1
X
..-_ _ _ _ _---'_.. The velocity of the particle..
The path OM will be the arc of a circle whose centre is at Q.2
ELECTROMAGNETIC DEFLECTION SENSITIVITY
Electronic Devices and Circuits
Magnetic Deflection in a Cathode Ray Tube
Electron will get deflected in magnetic field also.. . So a Cathode Ray Oscilloscope (CRO) can employ magnetic field as well to get deflective on the screen... In the region of the uniform magnetic field.32
1.. Let us derive the expression for the deflection sensitivity Sm in a magnetic field.. since 8 is small.5... The electron moves in a straight line from the cathode to the boundaries '0' of the magnetic field.20 Magnetic deflection sensitivity..

5 cm
(a)
For
Va = 500 V
Deflection Sensitivity SE= (b) (c) For For
2x~.Y.Vo
I.072 cmN
Va = 1000 V.11
In a CRT.. SE = 0.= = R mv m~2emVo
8= .eB l. Calculate the deflection sensitivity in m/volt if the final anode voltage is (a) 500 V (b) 1000 V (c) 1500 V
Solution
ID Deflection sensitivity = 2sVa
1=2 cm
D= 18 cm.B
V. it is used in radars and T.Electron Dynamics and CRO
33
But radius of the circle is mv R=eB In all practical cases.e. SMB
Advantages
I I Compared to electrostatic deflection SM is high.
~2Vo
~2. it will pass through the centre 0' of the region of the magnetic field. and since SE a -V whereas SM a ~.= . tubes. Va = 1500 V.= . o "Vo Because of high deflection sensitivity..
Disadvantages
For high frequency this is not used because of the high reactance produced by the coils.
SE = 0.
Ie
y=L
V.. therefore if MP' is projected backwards.
Ie
Deflection sensitivity
-1. the length of the deflecting plates in the direction of the beam is 2 cm.
.5 cm and the distance of the fluorescent screen from the centre of the plate is 18 cm. Radial deflection can be accomplished easily by rotating coils placed outside.036 cmN..
Problem 1.
s = 0. the spacing ofthe plates is 0. y=L tan 8 when 8 is small.B But 8= . L » I.~~8500
=
0. Ion spot formation creates a black spot at the centre of the screen.B
I.024 cmN. to produce B.. y = L8 I l.

. 7...... ...... The acceleration of an electron placed in an electric field of intensity & is a = ............... An electron takes T seconds to travel the distance L......................... The trajectory of an electron in two dimensional motion subjected to an electric field between two plates of a capacitor is .....
8. ................ Expression for the radius of an electron placed in a magnetic field of intensity B.. ....... 14....................
24. deflection is independent of specific charge (elm).... When suitable alternating voltages are impressed against two pairs of deflecting p!ates in a CRO................ The mass of a particle moving with a velocity v in terms of its rest mass mo is given by m = .... The trajectory of an electron moving with velocity v in a magnetic field B is The time taken T for one revolution of an electron in a magnetic field............ the magnetic force results in .................... ........... The total number of electrons passing through any cross section in unit times is .................................... 9... 20........... deflection sensitivity is inversely proportional to the square of accelerating potential........... figures are obtained on the screen. 5...
15........... .. 11... motion of the electron........... .. Electromagnetic deflection sensitivity is defined as .... ...........
12.. Its units are ........... volts.. the force experienced by the conductor is F m = ....... The ........
23..... .. 13.......
19................ Electric Field Intensity E = ................... Coating which provides the return path for electrons after striking the CRO Screen Cathode Material used in Cathode Ray Tube .
21..................................... .................. the famous ..... When a current carrying conductor oflength L is placed in a magnetic field of strength B..... ............... 10........ ........... ....... Electrostatic deflection sensitivity is defined as ........... fm = ........ The nature of signal ( wave shape) of Time Base in a CRO is ........... Expression for the velocity of electron v in terms of acclerating potential V = ... Force experienced by an electron in a magnetic field of Intensity B. .........
18. ... The .... 2........
.Electron Dynamics and CRO
37
OBJECTIVE TYPE QUESTIONS
1........... ............. Newtons.............. The energy acquired by an electron in rising through a potential of one volt is A Conductor of length L consists of N electrons....... 4........ 25...... .......
22......... Expression for electrostatic deflection sensitivity SE = .......................................
17...............
3..... Graduated scales on the CRO Screen are called as .................. When an electron enters the magnetic field with a velocity v in a direction perpendicular to the magnetic field......... and moving with velocity v is r = ........... Relativistic correction should be applied if an electron falls through a potential of above .......... is T = ..... 6........ 16.. Expression for electromagnetic deflection sensitivity SM = ........

(a) 5. volts the final velocity is. in time t secs.
Zero
'&' vim and magnetic
B£ em
(b)
Be m
e.6 x 10. With the help of a neat sketch. The force experienced by an electron in parallel electric field field 'B' wb/m1.
£
(c)
B rrt
(d)
5. If the electron starts at rest with initial velocity potential +V..JV mlsec
5.19 (d) 1. 7.59 x 10 .
2.. of charge e.
3. Give the constructional details of Cathode Ray Tube.6 x 10 19 Joules
2. and if the initial velocity is perpendicular to magnetic field IS. describe principle and working of Cathode Ray Tube. the path of the electron is
(a) (c) Common parabola Prolate parabola (b) (d) Curtate parabola Trochoid
6. 6.19
1. Compare Electrostatic and Electromagnetic deflection mechanisms What are the applications of CRO? Give the block schematic of CRO and explain.7 x 10-9 (c) 6. Derive the expression for electrostatic deflection sensitivity Derive the expression for electromagnetic deflection sensitivity. when placed in an accelerating field &..JV mlsec 5 10 .
4.JV cmlsec 3.. CRT screens with medium persistance will have visible glow on the screen for a period of
(a) 1 minute (b) 2 secs (c) 1 m sec (d) few minutes
. 5.. Derive the expression for the velocity acquired by an electron.
Joules
(b) 1.53 x
=0
x
m/Sec and is accelerated by
(c)
.1 x 10. Derive the expression for radius 'R' and time period 'T' in the case of an electron placed in a magnetic field of intensity B tesla. with electron density N per mJ . 1 ev
(a)
= .JV Km/sec
Ne tL
(b)
(d)
.. 8. is .38
Electronic Devices and Circuits ESSAY TYPE QUESTIONS
1. If aL electron is placed in combined '&' and 'B' fields and if '&' and 'B' are perpe ldicular to each other.
MULTIPLE CHOICE QUESTIONS 1. The expression for current I passing through a conductor of cross sectional area 'A'm1. with initial velocity zero m/sec.93)( 105 9.9 105
5
3. is 1=
Ne (a)
-t-
(b)
(c)
Ne tA
(d)
NA et
4.

Atomic Structure. Concept of Hole. transport mechanism.
• The basic aspects connected with Semiconductor Physics and the terms like Effective Mass. After p-type and n-type semiconductors.In this Chapter. Configurations. Intrinsic. Conductivity inp-type and n-type semiconductors are given . tunnel diode and so on. We shall also study the physical phenomena such as corniuction. This forms the basis to study Semiconductor Devices in the following chapters. and applications of semiconductor diodes. and Semiconductors are introduced. we shall study semiconductor devices formed using these two types of semiconductors.
•
•
•
. electrical characteristics. zener diode. Extrinsic.

there is one positively charged nucleus ( a proton) and a single electron. The force of attraction between the electron and the nucleus is : F a e2 . So hydrogen atom is neutral in charge.1
REVIEW OF SEMICONDUCTOR PHYSICS
ENERGY LEVELS AND ENERGY BANDS
To explain the phenomenon associated with conduction in metals and semiconductors and the emission of electrons from the surface of a metal. Rutherford found that atom consists of a nucleus with electrons rotating around it.
361t
As the electron is moving around the nucleus in a circular orbit with radius r. in terms of total energy oW' of the electron.)
r
v2
.1 2. a = . In hydrogen atom.F/m. The proton in the nucleus carries the charge of the atom.1.
2. The force of attraction between electron and proton follows Coulombs Law. we shall first study the basic aspects of Atomic Theory. Electronic structure of Silicon and Germanium elements. The mass of the atom is concentrated in the nucleus. Assume that the orbit of the electron around nucleus is a circle. and then conduction in Semiconductors. (Therefore. The electron will be moving around it in a closed orbit. Energy Band Theory. Its value is . We want to calculate the radius of this circle. m is the mass of electron and acceleration.. then the force of attraction given by the above expression F should be equal to the
2
centripetal force mv
r
(according to Newton's Second Law of Motion)
( F = m. It consist of protons which are positively charged.a. the nucleus has the proton with electron charge equal to 'e')
F a 1
~
( r is the radius of the orbit)
where
Eo
10-9 is the permitivity of free space. so it is immobile. {directly proportional to product of charges and inversely proportional to (distance )2}.40
Electronic Devices and Circuits
In this chapter. The charge on particle is positive and is equal to that of electron. we have to assume that atoms have loosely bound electrons which can be removed from it. and with velocity v. Semiconductor Device Physics.

. Therefore the energy of the electron is negative.. (Any quantity less than 0 is negative). then its total energy OW' must decrease by the amount equal to the radiation energy. But if the electron is radiating energy. So OW' should go on decreasing to satisfy the equation. Only for Hydrogen atom W will also be the energy of atom since it has only one electron.. Potential Energy is zero.-8-1tEor
e2
W is the energy ofthe electron. Therefore. r should decrease. So.... . But according to classical laws of electromagnetism.41tEor
= -2 mv --4-1tEor
I
2
e2
reduces to
e2 e2 e2 W= .81tEor 41tEor 81tE or
Energy possessed by the electron W
= .. Smce If .= .Junction Diode Characteristics
e2 e2 --4-1tEor
41
The Potential Energy of the electron
= ---2
41tEor
X
r
=
( -ve sign is because Potential Energy is by definition work done against the field) Kinetic Energy = ~ mv 2 Total Energy W possessed by the electron But e2 mv 2 = . The negative sign arises because the Potential Energy of the electron is
81tEor
As radius r increases. Potential Energy decreases. It is moving with velocity v or acceleration ~ around the nucleus.should decrease. So classical model of atom is not fairly satisfactory. Electron is having charge
= e. an accelerated charge must radiate energy.. this quantity
81tEor
should become more negative. then the frequency of the radiated energy should also be the same. If r is < 00. e2 If W decreases. The above equation is derived from the classical model of the electron. e2 W=--81tE o r
. r should also decrease. If the charge is performing oscillations with a frequency 'f'. When r is infinity. Hence the frequency of the radiated energy from the electron should be equal to the frequency with the electron orbiting round the nucleus. the energy should be less. the electron should describe smaller and smaller orbit and should finally fall into nucleous. Therefore this
r
2
electron should also radiate energy.

1. An electron moving through the I~ttice can be represented by a wave packet of plane waves grouped around the same value of K which is a wave vector. This is because the electron wave further )ecomes standing wave at the top and bottom ofa band i. While in states corresponding to these discrete energy levels.e. when the atom is in stationary state.1..2 THE BOHR ATOM
Electronic Devices and Circuits
The above difficulty was resolved by Bohr in 1913. Only during transition. The atom remains in the new state corresponding to WI. A stationary state is determined by the condition that the angular momentum of the electron in this state must be an integral multiple of h127r. They move under the combined influence of an external field plus that of a periodic potential of atom cores in the lattice.42 2. other than zero. i.e. When its energy changes from W2 to WI then the atom is said to have moved from one stationary state to the other.
nh 2tr
where n is an integer.. h = 6.
2. So mvr
=-
h
3. electron does not emit radiation in stationary state.sec. When the energy of the electron is changingfrom W2 to WI then radiation will be emitted The frequency of radiation is given by f= W2 -W\ where 'h' is Plank s Constant.
The atom can possess only discrete energies. He postulated three fundamental laws. E = Kinetic Energy of the Free Electron p=Momentum v=Velocity m = Mass p=mv m* = Effective mass of electron
p2
E=2m Electrons in a solid are not free. will some energy be radiated.626 x 10-34 J .
1.3
EFFECTIVE MAss
An electron mass 'm' when placed in a crystal lattice. Vg = O. The reason for this is the interaction of the electron even within lattice. responds to applied field as if it were of mass m*.
2.
. Electron velocity falls to zero at each band edge. it does not emit any radiation.